CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 16

Question. Determine K, so that the equation x² – 4x + k = 0 has no real roots.
Answer: K > 4

Question. Find the sum & product of the roots of the equation x² – = 0
Answer: sum = 0 , Product = – √13

Question. Find the equation whose roots are 5 + and 5 –
Answer: X² – 10x + 23 = 0

Question. Divide 51 into two parts whose product is 378.
Answer: 42,9

Question. Find the value of K for which the following quad. Equation have equal roots
(a). (k – 4) x² + 2(k – 4) x + 4 = 0
(b). kx (x – 2) + 6 = 0
Answer: (a) P ≤ 4 (b) P ≥ -8 (c) P ≥ 8, ≤ – 8

Question. If ∞ and β are the roots of the equation x² – 4x –5 = 0. Find ∞^-1 + β^-1
Answer: -4/5

Questions of 2 Mark

Question. Find two consecutive multiple of three whose product is 270
Answer: 15,18 ; -15, -18

Question. Solve the following by factorization method
(a) 2x² + ax – a² = 0, a ∈ R
(b) 4x² – 4ax + (a² – b²) = 0 (a,b ∈ R)
Answer: (a) x = -a , a/2 (b) a + b/2

Question. Is the following situation possible ? if so determine their present ages The sum of the ages of two friend are 20 years. 4 years ago, the product of their age in years was 48.
Answer:Not possible

Question. If one root of the quadratic equation 2x² + kx – 6 = 0 is 2, find the value of k also find the other root
Answer: 2, -3/2

Question. If the list price of a toy is reduced by Rs 2 a person can by 2 toys more for Rs 360.find original price of the toy.
Answer: Rs.20

Question. If – 5 is a root of 2x² + px – 15 = 0 & the quad. Equation f (x² + x) + k = 0 has equal root find the value of k
Answer: P = 7, k = 7/4

Question. Using quadratic formula, solve the following equation for x abx² + (b² – ac) x – bc = 0 (a,b,∈ R)
Answer: X = -b/a. x = c/b

Question. Two no. differ by 4 and their product is 192, find the numbers..
Answer: 12,16

Question. If α, β, be the roots of the equation ax² + bx + c = 0, then ax² + bx + c =.   
(A) (x – α)(x – β)
(B) a(x – α)(x – β)
(C) a (x – β) (x + α)
(D) a(x + α)(x + β)

Answer: C

Question.The number of real roots of the equation (x – 1)² + (x – 2)² + (x – 3)² = 0 is:   
(A) 2
(B) 1
(C) 0
(D) 3

Answer: C

Question.Determine k such that the quadratic equation x² + 7(3 +2k) – 2x(1 + 3k) = 0 has equal roots: 
(A) 2, 7
(B) 7, 5
(C) 2, -10/9
(D) None of these

Answer: C

Question. The condition that the equation ax² + bx + c = 0 has one positive and other negative root is:   
(A) a and c will have same sign
(B) a and c will have opposite signs
(C) b and c will have same sign
(D) b and c will have opposite signs

Answer: B

Question. The condition that both roots of the equation ax² + bx + c = 0 are negative is:   
(A) a, b, c are of the same sign
(B) a and b are of opposite signs
(C) b and c are of opposite signs
(D) the absolute term is zero

Answer: A

Question. If one root of 5x² + 13x + k = 0 is reciprocal of the other, then the value of k is. 
(A) 5
(B) 7
(C) 4
(D) None of these

Answer: A

Question. If the equation x²-bx / ax-c = m-1/m+1 has roots equal in magnitude but opposite in sign, then m is equal to.   
(A) a+b /a -b
(B)  a-b/a+b
(C) Both (A) & (B)
(D) None of these

Answer: B

Question. The set of values of p for which the roots of the equation 3x² + 2x + (p – 1)p = 0 are of opposite sign, is: 
(A) (0, 1)
(B) (-1,1)
(C) (-2,2)
(D) None of these

Answer: A

Question. Find the roots of the equation f(x) = (b – c) x² + (c – a) x + (a – b) = 0 
(A)  a-b /b-c and 1
(B) a+b /b+c and 1
(C) a-b /b+c and 1
(D) None of these

Answer: A

Question. The real values of a for which the quadratic equation 2x² – (a³ + 8a – 1)x + a² – 4a = 0 possesses roots of opposite signs are given by: 
(A) a > 6
(B) a > 9
(C) 0 < a < 4
(D) a < 0

Answer: C

Question. Given that (x + 1) is a factor of x² + ax + b and x² + cx – d, then   
(A) a + d = b + c
(B) a = b + c + d
(C) a + c = b - d
(D) None of these

Answer: B

Question. The zeroes of x² – bx + c are each decreased by 2 The resulting polynomial is x² – 2x + 1 Then   
(A) b = 6, c = 9
(B) b = 6, c = 3
(C) b = 3, c = 6
(D) b = 9, c = 6

Answer: A

Question. If the zeroes of x² + Px + t are two consecutive even numbers find the relation between P and t 
(A) P² – 4t + 4 = 0
(B) 4t – P² + 4 = 0
(C) - 4t² – 4 – P² = 0
(D) None of these

Answer: B

Question. One zero of x² – bx + C is the kth power of the other zero, then C ^1/k+1+C^k/k+1 is equal to 
(A) - b
(B) C
(C) - C
(D) b

Answer: D

Question. α, β, γ, δ are zeroes of x⁴ + 5x³ + 5x² + 5x – 6, then find the value of 1/α + 1/β + 1/γ +1/δ   
(A) 5/6
(B) -6/5
(C) -5/6
(D) None of these

Answer: A

Question. The number of real solution of the equation 2^{3x2-7x + 4} = 1 is: 
(A) 0
(B) 4
(C) 2
(D) Infinitely many

Answer: C

Question. The maximum value of – 3x² + 4x – 5 is at x =   
(A) 2/3
(B) 1/3
(C) -33/9
(D) None of these

Answer: A

Question. Given that 7 – 3i is a zero of x² + px + q, find the value of 3q + 4p 
(A) 14
(B) 58
(C) 118
(D) - 14

Answer: C

Question. Given that α is a zero of x⁴ + x² – 1, find the value of (α⁶ + 2α⁴)1000.   
(A) 1
(B) 0
(C) Either 0 or 1
(D) None of these

Answer: A

VERY SHORT ANSWER TYPE QUESTIONS

Question. If one root of the equation 3x² + px + 4 = 0 is 2/3 , find the value of p.

Answer: p = -8

Question. Determine whether x = - 1/3 and x = 2/3 are the roots of 9x² – 3x – 2 = 0.

Answer: yes

Question. If sum of a whole number and its reciprocal is 10/3 , what is the number?

Answer: 3

Question. For what value of k, x = a is a solution of the equation x² - (a + b)x + k = 0 .

Answer: k = ab

Question. Find the ratio of sum and product of the roots of the equation 3x² – 8x + 12 = 0.

Answer: 2 : 3

Question. Form a quadratic equation whose roots are –2 and 3.

Answer: x² - x - 6 = 0

Question. If one root of a quadratic equation is 3 - √2/4 , then what is the other root?

Answer: 3 + √2/4

Question. Form a quadratic equation whose sum of roots is –3 and product of roots is 5.

Answer: x² + 3x + 5 = 0

Question Write the quadratic equation for the following : The sum of two numbers is 27 and their product is 182.

Answer: x² - 27x +182 = 0

Question. The sum of squares of two consecutive positive integers is 365. Express this in the form of an quadratic equation.

Answer: x² + x – 182 = 0

Question. Represent the following in the form of a quadratic equation : The sum of the ages of father and his son is 45 years. Five years ago, the product of their ages (in years) was 124.

Answer: x² - 45x + 324 = 0

Question. Write the discriminant of the equation 6a²x² – 7abx – 3b² = 0.

Answer: 121a²b²

Question. Write the nature of the roots of the equation 3x² - 4 √3x - 1 = 0.

Answer: Real and distinct roots

Question. Check whether x (x + 3) + 8x = (x + 5) (x – 5) is a quadratic equation.

Answer: No

Question. For what value of k, the equation 3x² + kx + 4 = 0 have equal roots?

Answer: k = ±4√3

Question. If x = 3/2 is the root of the quadratic equation 2x² – kx + 3 = 0, then find the value of k.

Answer: k = 5

Question. For what value of k, the equation kx² + 6x + 1 = 0 have real roots?

Answer: k ≤ 9

Question. If α and β are the roots of the equation 3x² – 9x + 7 = 0, then find α + β + αβ.

Answer: 16/3

Question. What are the roots of the equation 2x² + 3x – 2 = 0?

Answer: -2 , 1/2

Question. Solve the quadratic equation : (x + 3)² – 16 = 0.

Answer: 1, –7

Short Answer type Question :

Question. Find the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other. 
Solution. Let the roots of the given equation be a and 1/α 
α , 1/α = c/a = k/3 ⇒ k = 3

Question. Find the value of k so that the quadratic equation kx(3x – 10) + 25 = 0, has two equal roots.
Solution. kx(3x – 10) + 25 = 0
⇒ 3kx² – 10kx + 25 = 0
D = (–10k)² – 4 × 3k × 25
= 100k² – 300k
For equal roots, D = 0 (1)
⇒ 100k² – 300k = 0
⇒ 100k(k – 3) = 0
⇒ k = 0 or k = 3
But k π 0,
So, k = 0 (Rejected)
[∴ In quadratic equation, a ≠ 0]
Hence, k = 3

Question. For what values of k, the equation 9x² + 6kx + 4 = 0 has equal roots? 
Solution. 9x² + 6kx + 4 = 0
(6k)² – 4 × 9 × 4 = 0 (½)
36k² = 144
⇒ k2 = 4
k = ±2

Question. Find the value(s) of k so that the quadratic equation x² – 4kx + k = 0 has equal roots.
Solution. x² – 4kx + k = 0
Since given equation has equal roots,
∴ D = 0 (1)
16k² – 4k = 0
⇒ 4k(4k – 1) = 0
⇒ k = 0 and k = 1/4

Question. For what value(s) of ‘a’ quadratic equation 3ax² – 6x + 1 = 0 has no real roots?
Solution. 3ax² – 6x + 1 = 0 (½)
(–6)² – 4(3a)(1) < 0
12a > 36
⇒ a > 3

Question. State whether the quadratic equation 4x² – 5x + 25/16 = 0 has two distinct real roots or not. Justify your answer.
Solution. No, D = 0

Question. Find the value(s) of k so that the quadratic equation 3x² – 2kx + 12 = 0 has equal roots. 
Solution. 3x² – 2kx + 12 = 0
Since given equation has equal roots, so
D = 0 (1)
⇒ 4k² – 144 = 0
⇒ 4(k² – 36) = 0
⇒ k = ±6
⇒ k = 6 and k = –6

Question. Find the value of p, so that the quadratic equation px(x – 3) + 9 = 0 has equal roots. 
Solution. px(x – 3) + 9 = 0
⇒ px² – 3px + 9 = 0
When roots are equal,
D = b² – 4ac = 0
9p² – 36p = 0
⇒ 9p(p – 4) = 0
⇒ p = 0, p = 4
But p π 0
[∴ In quadratic equation, a ≠ 0]
∴ p = 4

Question. Find the values of k for which the quadratic equation 9x² – 3kx + k = 0 has equal roots.
Solution. For equal roots, D = 0
⇒ 9k² – 36k = 0
⇒ 9k(k – 4) = 0 
⇒ k = 0 or k = 4

Question. Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0, hence find the nature of its roots. 
Solution. –8, no real roots

Question. If x = 3 is one root of the quadratic equation x² – 2kx– 6 = 0, then find the value of k. 
Solution. x = 3 is one root of the equation
∴ 9 – 6 k – 6 = 0
⇒ k = 1/2

Question. Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also find these roots.
Solution.  For equal roots, D = 0
{2(k + 1)}2 – 4(3k + 1) · 1 = 0 (1)
⇒ 4(k² + 2k + 1) – 12k – 4 = 0
⇒ 4k² + 8k + 4 – 12k – 4 = 0 (1)
⇒ 4k² – 4k = 0
⇒ 4k(k – 1) = 0
⇒ k = 0, 1

Question. Find the value(s) of k so that the quadratic equation 2x² + kx + 3 = 0 has equal roots. 
Solution. 2x² + kx + 3 = 0
For equal roots, D = 0 
⇒ b² – 4ac = 0
⇒ k² – 24 = 0
⇒ k = ± 2√6 

Question. If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k. 
Solution. 3(2)² + p(2) – 8 = 0
⇒ 12 + 2p – 8 = 0
⇒ p = – 2 ...(i)(1)
So, equation becomes
4x² + 4x + k = 0
For equal roots, D = 0
⇒ (4)² – 4 × 4 × k = 0
⇒ 16 = 16k
⇒ k = 1

Question. Find the value of p for which the quadratic equation (p + 1)x² – 6(p + 1)x + 3(p + 9) = 0, p π –1 has equal roots. Hence, find the roots of the equation. 
Solution. 3, 3.

Question. For what values of k, the roots of the equation x² + 4x + k = 0 are real? 
Solution. For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ (4)² – 4 × 1 × k ≥ 0
⇒ 16 – 4k ≥ 0
⇒ 16 ≥ 4k, k ≤ 4

Question. Find that non-zero value of k, for which the quadratic equation kx² + 1 – 2(k – 1)x + x² = 0 has equal roots. Hence, find the roots of the equation.
Solution. kx² + 1 – 2(k – 1)x + x² = 0
⇒ (k + 1)x² – 2(k – 1)x + 1 = 0
Q Above equation has equal roots, 
So, discriminant, D = 0
⇒ {–2(k – 1)}2 – 4 × (k + 1) × 1 = 0 
⇒ 4(k² – 2k + 1) – 4(k + 1) = 0
⇒ 4k² – 12k = 0
⇒ 4k(k – 3) = 0
⇒ k = 3 (as k ≠ 0) 

Question. Find whether the equation 1/2x - 3 + 1/x - 5 = 1 , x ≠ 3/2 , 5 has real roots. If real roots exist, find them.
Solution. Yes, 8 ±3√2/2

Question. The roots a and b of the quadratic equation x² – 5x + 3(k – 1) = 0 are such that a – b = 1. Find the value k.
Solution. k = 3

Question. Find the values of k for which the given equation has real and equal roots: (k + 1)x² -2(k - 1)x + 1 = 0 
Solution. We have, (k+1)x² - 2(k - 1)x+1 = 0.
a = k + 1, b = -2(k - 1), c = 1.
D = b² - 4ac =4(k-1)² - 4(k + 1) =4(k² -3k)
The given equation will have real and equal roots, if
D = 0 ⇒ 4 (k² - 3k) = 0 ⇒ k² - 3k = 0 ⇒ k (k - 3) = 0 ⇒ k = 0, 3

Question. Solve the quadratic equations by factorization method: x² - 9 = 0
Solution. We have,
x² - 9 = 0
⇒ (x - 3)(x + 3) = 0
⇒ x - 3 = 0 or, x + 3 = 0
x = 3 or, x = -3 ⇒ x = ± 3
Thus, x = 3 and x = - 3 are roots of the given equation.

Question. If p, q, r and s are real numbers such that pr = 2(q + s), then show that at least one of the equations x² + px + q = 0 and x² + rx + s = 0 has real roots. 
Solution. Given quadratic equations are;
x² + px + q = 0 —(i)
and, x² + rx + s = 0 ......(ii)
Also given ; pr = 2(q + s)........(iii)
Let D₁ and D₂ be the discriminant of quadratic equations (i) and (ii) respectively.
Then,
D₁ = p² - 4q and D² = r2 - 4s
⇒ D₁+ D₂ = p² - 4q + r² - 4s = (p² + r²) - 4(q + s)
Now, Since sum of both D₂ & D₁ is greater than or equal to 0. Hence, both can't be
negative.
⇒ At least one of D₁and D₂ is greater than or equal to zero
Case 1. If D₁ ≥ 0, equation (i) has real roots.
Case 2.If D₂ ≥ 0, equation (ii) has real roots.
Case 3. If D₁ & D₂ both ≥ 0, then equation (i) & (ii) both have equal roots.
Clearly, from case 1,2 & 3 at least one given quadratic equations has equal roots.

Question. Check whether the given equation is quadratic equation: (x-3) (2x + 1) = x(x + 5) 
Solution. The given equation is (x - 3) (2x +1) = x (x+5)
⇒ 2x² + x - 6x - 3 = x² + 5x
⇒ 2x² - 5x - 3 = x² + 5x
⇒ x² - 10x - 3 = 
It is in the form of ax² + bx + c = 0, a≠0
∴ the given equation is a quadratic equation.

Question. Form a quadratic equation whose roots are -3 and 4. 
Solution. We have, x = 4 and x = -3.
Then,
 x - 4 = 0 and x + 3 = 0
⇒ (x - 4)(x + 3) = 0
⇒ x² + 3x - 4x - 12 = 0
⇒ x² - x - 12 = 0
This is the required quadratic equation

Question. Check whether the given equation is quadratic equation: (x +1)²= 2 (x –3) 
Solution. The given equation is (x+1)² = 2 (x-3)
⇒ x² + 2x + 1 - 2x + 6 = 0
⇒ x² + 7 = 0
⇒ x² + 0.x + 7 = 0
Which is of the form ax² + bx + c = 0
Hence, the given equation is a quadratic equation.

Question. Find discriminant of the quadratic equation: 5x² + 5x + 6 = 0. 
Solution. Given equation is 5x² + 5x + 6 = 0
Here a = 5, b = 5, c = 6
D = b² - 4ac = (5)² -4 x 5 x 6 = -95

Question. Two numbers differ by 3 and their product is 504. Find the numbers. 
Solution. Sol : Let the required number be x and x + 3.
Then, according to given question we have,
x (x + 3) = 504
⇒ x² + 3x = 504
⇒ x² + 3x - 504 = 0
⇒ x² + 24x - 21x - 504 = 0
⇒ x(x + 24) - 21(x + 24) = 0
⇒ (x + 24)(x - 21) = 0
⇒ x + 24 = 0 or x - 21 = 0
⇒ x = - 24 or x = 21
Case I: When x = -24
∴ x + 3 = -24 + 3 = -21
Case II: When x = 21
∴ x + 3 = 21 + 3 = 24
Hence, the numbers are -21, -24 or 21, 24.

Question. The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers. 
Solution. Let the smaller number be x and the larger number be y.
Also, Square of the larger number( y² ) =18x
According to question,
x² + y² = 208
⇒ x² + 18x = 208
⇒ x² + 18x - 208 = 0
⇒ x² + 26x - 8x - 208 = 0
⇒ (x + 26) (x - 8) = 0 ⇒ x = 8, x = -26
But, the numbers are positive. Therefore, x =8
Square of the larger number = 18x = 18 x 8 = 144
Therefore, larger number =√144 = 12
Hence, the numbers are 8 and 12.

Question. The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age. 
Solution. Let the present age of tanvy be x years
Tanvy's age five years ago = (x - 5)
Tanvy's age eight years later = (x + 8)
According to question,
⇒ (x - 5)(x + 8) = 30
⇒ x² + 8x - 5x - 40 = 30
⇒ x² + 3x - 40 - 30 = 0
⇒ x² + 3x - 70 = 0
⇒ x² + 10x - 7x - 70 = 0
⇒ x(x + 10) - 7(x + 10) = 0
⇒ x + 10 = 0 or x - 7 = 0
⇒ x = -10 or x = 7
⇒ x = 7 ( age cannot be negative)
Therefore, the present age of tanvy is 7 years.

Question. Solve: 4x² - 12x + 9 = 0.
Solution. We have,
4x - 12x + 9 = 0
Here, 4 x 9 = 36 so to factor the middle term in given equation we have (-6) x (-6) = 36, and (-6) + (-6) = -12.
⇒ 4x² - 6x - 6x + 9 = 0 ⇒ 2x(2x - 3) - 3(2x - 3) = 0
⇒ (2x - 3)(2x - 3) = 0 ⇒ (2x - 3)2 = 0
2x - 3 = 0 ⇒ x = 3/2
Hence, x = 3/2 is the repeated root of the given equation.

Question. If –4 is a root of the quadratic equation x² + px – 4 = 0 and the quadratic equation x² + px + k = 0 has equal roots. Find the value of k. 
Solution. We have, x² + px - 4 = 0
-4 is the root of the given equation
Substitute x = - 4 in the given equation, we get
(-4)² + p (-4) -4 =0
⇒ 16 - 4p - 4 = 0
⇒ 4p = 12 or p = 3
The equation becomes x2 + 3x -4 = 0
Substituting the value of p = 3 in the equation x2 + px + k = 0, we get
x² + 3x + k = 0
Here a = 1, b = 3, c= k
∴ D = b² - 4ac = (3)2 - 4 (1) (k)
= 9 - 4k
For equal roots, D = 0
⇒ 9 - 4k = 0 or k = 9/4

Question. If x = - 2 is a root of the equation 3x² + 7x + p = 0, find the values of k so that the roots of the equation x² + k(4x + k - 1) + p = 0 are equal. 
Solution. Here, x = - 2 is a root of 3x² + 7x + p = 0
⇒ 3(-2)2 + 7 (-2) + p = 0
⇒ p = 2
∴ x² + k(4x + k - 1) + p = 0 becomes
x² + 4kx + k² - k + 2 = 0....(i)
Comparing eq. (i) with ax² + bx + c = 0 , we get
a = 1 , b = 4k and c = k² - k + 2
Roots of eq (i) are equal
So, D =b² - 4ac =0
⇒ (4k)² - 4 x 1 (k² - k + 2) = 0
⇒ 16k² - 4k² + 4k - 8 = 0
⇒ 12k² + 4k - 8 = 0
⇒ 12k² + 4k - 8 = 0
⇒ 3k² + k - 2 = 0
⇒ 3k² + 3k - 2k - 2 = 0
⇒ 3k(k + 1) - 2(k + 1) = 0
⇒ (k + 1)(3k - 2) = 0
⇒ k = -1, k = 2/3

Question. Find the discriminant of equation: 2x² - 7x + 6 = 0. 
Solution. Given, 2x² - 7x + 6 = 0
∴ a =2, b = -7 and c = 6
D = b² -4ac
= (-7)2 - 4(2)(6)
= 49 - 48
= 1

Question. Solve the following problem: x² − 45x + 324 = 0 
Solution. x² - 45x + 324 = 0
⇒ x² - 36x - 9x + 324 = 0 ⇒ x (x - 36) - 9(x - 36)=0
⇒ (x - 9)(x - 36) ⇒ x = 9, 36

Question. At t minutes past 2 p.m, the time needed by the minute hand of a clock to show 3 p.m. was found to be 3 minutes less than t²/4 minutes. Find t. 
Solution. Total time taken by minute hand from 2 p.m. to 3 p.m. is 60 min.
According to question,
t + (t²/4 - 3) = 60
⇒ 4t + t² – 12 = 240
⇒ t² + 4t – 252 = 0
⇒ t² + 18t – 14t – 252 = 0
⇒ t(t + 18) – 14(t +18) = 0
⇒ (t + 18) (t – 14) = 0
⇒ t + 18 = 0 or t – 14 = 0
⇒ t = –18 or t = 14 min.
As time can't be negative. Therefore, t = 14 min.

Question. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.
Solution. Let the ten's place digit be y and unit's place be x.
Therefore, number is 10y + x.
According to given condition,
10y + x = 4(x + y) and 10y + x = 2xy
⇒ x = 2y and 10y + x = 2xy
Putting x = 2y in 10y + x = 2xy
10y + 2y = 2.2y.y
12y = 4y²
4y² - 12y = 0 ⇒ 4y(y - 3) = 0
⇒ y - 3 = 0 or y = 3
Hence, the ten's place digit is 3 and units digit is 6 (2y = x)
Hence the required number is 36.

Question. Check whether it is quadratic equation: (x + 1)³ = x³ + x + 6 
Solution. We have the following equation,
(x + 1)³ = x³ + x + 6
⇒ x³ + 1 + 3x(x + 1) = x³ + x + 6
⇒ 3x² + 2x - 5 = 0.
This is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

Question. Find the roots of the quadratic equation 2x² - x - 6 = 0 
Solution. Given, 2x² - x - 6 = 0
Splitting the middle term of the equation,
⇒ 2x² - 4x + 3x - 6 = 0
⇒ 2x(x - 2) + 3(x - 2) = 0
⇒ (x - 2)(2x + 3) = 0
⇒ x - 2 = 0 or 2x + 3 = 0
Therefore, x = 2 or x = -3/2

Question. A journey of 192 km from a town A to town B takes 2 hours more by an ordinary passenger train than a super fast train. If the speed of the faster train is 16 km/h more, find the speed of the faster and the passenger train. 
Solution.
or x(x + 48) - 32(x + 48) = 0
or, (x - 32) (x + 48) = 0
or, x = 32 or - 48
Since speed can't be negative, therefore - 48 is not possible.
∴ Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h

Question. Solve the following problem: x²−55x +750=0
Solution. x ²−55x +750=0
⇒ x ²−25x−30x +750=0 ⇒ x (x−25)- 30(x−25)=0
⇒ (x−30)(x−25) ⇒ x =30, 25

Question. Check whether(x - 7)x = 3x² - 5 is a quadratic equation:   
Solution. Given equation is (x - 7) x = 3x² - 5
⇒ x² - 7x = 3x2 -5
⇒ 2x² + 7x - 5 = 0
which of the form ax² + bx + c = 0
Hence, given equation is a quadratic equation.

Question. What is the nature of roots of the quadratic equation 5x² - 2x - 3 = 0? 
Solution. On comparing equation with standard form of equation i.e, ax² + bx + c = 0, we get
a = 5, b = -2, c = -3
Now , D = b² - 4ac = (-2)2 - 4 x 5 x (-3)
= 4 + 60 = 64
Therefore, D = 64
We know , For D>0 , the roots of equation are real and distinct.
Therefore , 5x² - 2x - 3 = 0 has real and distinct roots.

Question. A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side-fence. Find the dimensions of his garden.
Solution. Let the length of one side be x metres and other side be y metres.
Then, x + y + x = 30 y = 30 ⇒ - 2x
Area of the garden = 100 m2
⇒ xy = 100
⇒ x (30-2x) = 100
⇒ 2x² - 30x +100 = 0
⇒ 2(x²-15x+50) = 0 or x²-15x+50 = 0
⇒ x²-10x - 5x +50 = 0
⇒ x (x-10) - 5 (x-10) = 0
⇒ (x-10) (x-5) = 0
Either x-10 = 0 or x-5 = 0
⇒ x= 10, 5
∴ y = 30 - 20 = 10 or 30 -10 =20
Hence, the dimensions of the vegetable garden are 5m 20m or 10m 10m.

Question. A faster train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.
Solution. 40 km/hr, 50 km/hr

Question. X and Y are centres of circle of radius 9 cm and 2 cm and XY = 17 cm. Z is the centre of a circle of radius r cm, which touches the above circles externally. Given that ∠XZY = 90°, find the value of r.
Solution. r = 6 cm

Question. A two digit number is such that the product of its digit is 10. When 27 is subtracted from the number, the digits interchange their places. Find the number.
Solution. 52

Question. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution. 36

Question. Find the values of k so that the equation (k + 2) x² – (7 – k) x + 3 = 0 has :
(a) equal roots (b) distinct roots (c) no real roots.
Solution. (a) k = 1, 25 (b) k > 25 or k < 1 (c) 1 < k < 2

Question. The sum of the ages of a man and his son is 40 years. The product of their ages is 144 years. Find their present ages.
Solution. 36 years and 4 years

Question. The sum of two numbers is 16. The sum of their reciprocals is 1/3 Find the numbers.
Solution. 4, 12

Question. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of the flight.
Solution. 1 hour

Question. The length of a rectangular ground is greater than twice its breadth by 5 m. The area of the ground is 1125 sq. m. Find the length and breadth of the ground.
Solution. length = 50 m, breadth = 45/2 m

Question. The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. If the area of the triangle be 60 sq. cm., calculate the length of the sides of the triangle.
Solution. 15 cm, 8 cm, 17 cm

Question. The difference of the squares of two numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
Solution. 6 and 9

Question. A carpet is placed in a room 6m × 4m, leaving a border of a uniform width all around it. Find the width of the border, if the area of the carpet is 8 sq. m.
Solution. 1 m

Question. Two circles touch externally. The sum of their areas is 130 π sq.cm and the distance between their centres is 14 cm. Find the radii of the circles.
Solution. 11 cm and 3 cm

Question. The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 sq. m. Find its dimensions.
Solution. 25 m × 20 m

Question. If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2. However, if twice th area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Find the sides of the squares.
Solution. 5 cm and 8 cm

Question. In an auditorium, seats are arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row is reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
Solution. (i) 30 (ii) 1200

Question. The area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm. Find its base.
Solution. 10 cm or 24 cm

Question. The sum of the squares of two consecutive odd positive integers is 290. Find them.
Solution. 11, 13

Question. A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. How many books did he buy?
Solution. 16

Question. A line AB is 8 cm in length. AB is produced to P such that BP² = AB. AP. Find the length of BP.
Solution. 4 (1 + √5) cm

Question. If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy. 
Solution. Rs. 20

Question. Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Solution. Swati’s present age = 9 years, varun’s present age = 27 years

Question. A person on tour has Rs. 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by Rs. 3. Find the original duration of the tour.
Solution. 20 days

Question. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hour 30 minutes. Find the speed of the stream.
Solution. 5 km/hr

Question. Rs. 6500 were divided among a certain number of persons. Had there been 15 more persons, each would have got Rs. 30 less. Find the original number of the persons.
Solution. 50

Question. A piece of cloth costs Rs. 200. If the piece was 5 m longer and each metre of cloth costs Rs. 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
Solution. 20 m, Rs. 10

Question. The denominator of a fraction is 5 more than its numerator. The sum of the fraction and its reciprocal is 3 , 11/14 .Find the fraction.
Solution. 2/7

Question. A takes 15 days less than the time taken by B to finish piece of work. If both A and B together can finish the work in 18 days, find the time taken by B to finish the work.
Solution. 45 days

Question. If the sum of first n even natural numbers is 420, find the value of n.
Solution. n = 20

Question. A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Solution. 25 km/hr

Question. Two pipes running together can fill a cistern in 3 , 1/3 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Solution. 5 minutes and 8 minutes

Question. The speed of a boat in still water is 8 km/hr. It can go 15 km upstream 22 km downstream in 5 hours. Find the speed of the stream.
Solution. 3 km/hr

Question. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution. 10, 6

Question. In a class test, the sum of Nishu’s marks in Mathematics and Science is 30. Had she got 2 marks more in Mathematics and 3 marks less in Science, the product of their marks would have been 210. Find her marks in the two subjects.
Solution. Marks in mathematics = 13, Marks in science = 17 OR Marks in Mathematics = 12, Marks in science = 18

Question. Find two consecutive numbers whose squares have the sum 85.
Solution. 6, 7

Question. A polygon of n side has n(n - 3) diagonals. How many sides has a polygon with 54 diagonals?
Solution. 12

Question. Two numbers differ by 3 and their product is 504. Find the numbers.
Solution. 21, 24 or – 24, – 21 

Question. A farmer wishes to start a 300 sq. m rectangular vegetable garden. Since he has only 50 m barbed wire, he fences three sides of the rectangular garden, letting his house compound wall act as the fourth side fence. Find the dimensions of the garden.
Solution. 20 m × 15 m or 30 m × 10 m

Question. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his sons age. Find their present ages.
Solution. 7 years and 49 years

Question. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
Solution. 150 cm²

Question. Sum of the areas of two squares is 468 m2. If the difference of their perimeter is 24 m, find the sides of the two squares.
Solution. 18 m and 12 m

Question. O girl! Out of a group of swans, 7/2 times the square root of the number are playing on the shore of a tank. The two remaining ones are playing with amorous fight, in the water. What is the total number of swans?
Solution. 16

Question. The angry Arjun carried some arrows for fighting with Bheesm. With half the arrows, he cut down the arrows thrown by Bheesm on him and with six other arrows, he killed the rath driver of the Bheesm. With one arrow each, he knocked down respectively the rath, flag and the bow of Bheesm. Finally, with one more than four times the square root of the total number of arrows, he laid Bheesm unconscious on an arrow bed. Find the total number of arrows Arjun had.
Solution. 100

Question. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If, after 2 hours, they are 50 km apart, find the speed of each trains.
Solution. 20 km/hr, 15 km/hr

Question. In an auditorium, the number of rows was equal to number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. How many rows were there?
Solution. 30 rows

Question. Some students planned a picnic. The budget the food was Rs. 500. But, 5 of them failed to go and thus the cost of food for each member increased by Rs. 5. How many students attended the picnic?
Solution. 20

Question. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km per hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train. 
Solution. 45 km/hr

Question. Out of a number of Saras birds, one fourth the number are moving about in lotus plants; 1/9 th coupled (along) with 1/4 th as well as 7 times the square root of the number move on a hill; 56 birds remain in vakula trees. What is the total number of birds?
Solution. 576

Question. A two digit number is such that the product of the digits is 14. When 45 is added to the number, then the digits are reversed. Find the number.
Solution. 27

Question. The length of the hypotenuse of a right-angled triangle exceeds the base by 1 cm and also exceeds twice the length of the altitude by 3 cm. Find the length of each side of the triangle.
Solution. 5 cm, 12 cm, 13 cm

Question. Find the values of k for which the equation x² + 5kx + 16 = 0 has no real roots.
Solution. - 8 /5 <k< 8/5

Question. Solve the equation 2x² – 5x + 3 = 0 by the method of completing the square.
Solution. 1 , 3/2

Question. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Solution. 4, 12

Question. A two digit number is four times the sum of its digits and twice the product of its digits. Find the number.
Solution. 36

Question. Solve for x using quadratic formula :
a²b²x² + b²x - a²x - 1 = 0
Solution. - 1/ a² , 1/b²

Question. Divide 26 into two parts, whose sum of squares is 346. Frame an equation for the given statement.
Solution. x² + (26 – x)² = 346

Question. Solve for x : 1/ x + 4 - 1/x - 7 = 11/30 ; x ≠ - 4, 7
Solution. 1, 2

Question. Solve for x : 4x² – 4ax + a² – b² = 0.
Solution. a + b /2 , x = a - b /2 

Question. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Solution. 800 km/hr

Question. Find the values of k so that the quadratic equation x² – 2x (1 + 3k) + 7 (3 + 2k) = 0 has equal roots.
Solution. 2 or - 10/9 

Question. Represent the following situation in the form a quadratic equation:
(a) Abdul and Sneha together have 30 oranges. Both of them ate 3 oranges each and the product of the number of oranges they have now is 120. We would like to find out how many oranges they had initially.
(b) The area of a rectangular plot is 428 m2. The length of the plot (in metres) is two more than twice its breadth. We need to find the length and breadth of the plot.
Solution. (a) Let the number of number of left oranges with Abdul and Sneha be x and y respectively.
Then, x + y = 30 ⇒ y = 30 – x
The number of oranges left with both Abdul and Sneha are x – 3 and y – 3 respectively.
The product of number of left oranges = 120
⇒ (x – 3) (y – 3) = 120
⇒ (x – 3) (30 – x – 3) = 120 (Q y = 30 – x)
⇒ (x – 3) (27 – x) = 120
⇒ 27x – x² – 81 + 3x = 120 ⇒ x² – 30x + 201 = 0
(b) Let the breadth of the plot be x. Then the length of the plot = 2x + 2
Since, area of the plot = 428 m² (Given)
∴ x(2x + 2) = 428
⇒ 2x² + 2x – 428 = 0
⇒ x² + x – 214 = 0

Question. Write the set of values of k for which the quadratic equation 2x² + kx + 8 has real roots. 
Solution. For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0 ⇒ k² – 4(2) (8) ≥ 0
⇒ k² – 64 ≥ 0 ⇒ k² ≥ 64 ⇒ k < –8 and k > 8

Question. Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.
Solution. Let the roots of the given equation be a and 6a.
Thus the quadratic equation is (x – a)(x – 6a) = 0
⇒ x² – 7ax + 6a² = 0 ...(i)
Given equation can be written as
x² - 14/P x + 8/P = 0
Comparing the coefficients in (i) and (ii) 7α= 14/P and 6α² = 8/P
Solving to get p = 3.

Question. State whether the equation (x + 1) (x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution. We have (x + 1) (x – 2) + x = 0 ⇒ x² – x – 2 + x = 0 ⇒ x² – 2 = 0
∴ D = b² – 4ac = 0 – 4(1) (–2) = 8 > 0
∴ Given equation has two distinct real roots.

Question 1. The value of k if \( x^2 - kx + 4 = 0 \) has equal roots. [Ans.: -4 or 4]
Answer: For any quadratic equation to have equal roots, its discriminant must be zero. We compare the given equation with the standard quadratic form \( ax^2 + bx + c = 0 \) to identify \( a = 1 \), \( b = -k \), and \( c = 4 \). Setting the discriminant \( D = b^2 - 4ac = 0 \): \[ (-k)^2 - 4(1)(4) = 0 \] \[ k^2 - 16 = 0 \] \[ k^2 = 16 \] Taking the square root on both sides: \[ k = 4 \text{ or } k = -4 \]
In simple words: A quadratic equation has equal roots when its discriminant is exactly zero. Solving this gives \( k = 4 \) or \( k = -4 \).

Exam Tip: Always specify both positive and negative values when taking square roots to avoid losing marks.

Question 2. If the Q.E \( ax^2 + bx + c = 0 \) has equal roots then each of the roots is _________. [Ans.: \( -\frac{b}{2a} \)]
Answer: When a quadratic equation \( ax^2 + bx + c = 0 \) has equal roots, its discriminant \( D = b^2 - 4ac \) is equal to zero. According to the quadratic formula, the roots are given by: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] Since the roots are equal, we substitute \( D = 0 \) into the formula: \[ x = \frac{-b \pm 0}{2a} = -\frac{b}{2a} \] Thus, each of the equal roots is represented by \( -\frac{b}{2a} \).
In simple words: When a quadratic equation has identical roots, the term inside the square root in the quadratic formula becomes zero, leaving both roots as \( -\frac{b}{2a} \).

Exam Tip: Memorize this formula as it is a common shortcut used in multiple-choice questions on equal roots.

Question 3. The equation \( 2x^2 + px + 3 = 0 \) has real roots then the value of p is _________. [Ans.: \( p \ge 2\sqrt{6}, p \le -2\sqrt{6} \)]
Answer: For the quadratic equation \( 2x^2 + px + 3 = 0 \) to have real roots, its discriminant \( D \) must be greater than or equal to zero. Here, \( a = 2 \), \( b = p \), and \( c = 3 \). We apply the inequality: \[ D = b^2 - 4ac \ge 0 \] \[ p^2 - 4(2)(3) \ge 0 \] \[ p^2 - 24 \ge 0 \] \[ p^2 \ge 24 \] Taking the square root on both sides yields: \[ p \ge 2\sqrt{6} \quad \text{or} \quad p \le -2\sqrt{6} \]
In simple words: For real roots, the discriminant must be at least zero. Solving \( p^2 - 24 \ge 0 \) gives the ranges for \( p \).

Exam Tip: Remember that inequalities split into two separate parts when taking the square root on both sides of a squared variable inequality like \( p^2 \ge 24 \).

Question 4. Without solving comment on the nature of the roots of the Quadratic Equation: \( 2x^2 + 10x + 49 = 0 \).
Answer: To determine the nature of the roots of the quadratic equation \( 2x^2 + 10x + 49 = 0 \) without solving it, we compute its discriminant \( D = b^2 - 4ac \). Here, \( a = 2 \), \( b = 10 \), and \( c = 49 \): \[ D = (10)^2 - 4(2)(49) \] \[ D = 100 - 392 = -292 \] Since the discriminant is less than zero (\( D < 0 \)), the given quadratic equation has no real roots. Therefore, the roots are imaginary.
In simple words: Since the discriminant is a negative number (-292), the equation has no real solutions.

Exam Tip: Clearly state that \( D < 0 \) implies no real roots to earn full marks on nature of roots questions.

Question 5. The sum of the roots of a Q.E \( ax^2 + bx + c = 0 \) is _________ and the product of the roots is _________
Answer: For any standard quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \), the sum and product of the roots are determined directly by its coefficients. The sum of the roots is given by the formula \( -\frac{b}{a} \), which is the negative ratio of the coefficient of \( x \) to the coefficient of \( x^2 \). Meanwhile, the product of the roots is calculated as \( \frac{c}{a} \), representing the ratio of the constant term to the coefficient of \( x^2 \).
In simple words: The sum of the roots is always the negative of the middle number divided by the first number, and the product is the last number divided by the first.

Exam Tip: Don't forget the negative sign in the sum formula \( -\frac{b}{a} \).

Question 6. If \( \alpha \) and \( \beta \) are the roots of the Q.E. \( ax^2 + bx + c = 0 \), then the Q.E. whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is _________.
Answer: Let the roots of the given quadratic equation \( ax^2 + bx + c = 0 \) be \( \alpha \) and \( \beta \). We know that: \[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a} \] For the new equation, the roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \). Let's find the new sum and product of roots: - New Sum \( (S') = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c} \) - New Product \( (P') = \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c} \) The new quadratic equation is given by: \[ x^2 - S'x + P' = 0 \] \[ x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} = 0 \] \[ x^2 + \frac{b}{c}x + \frac{a}{c} = 0 \] Multiplying by \( c \), we get the equation: \[ cx^2 + bx + a = 0 \]
In simple words: Swapping the first and last coefficients of the original equation creates a new equation whose roots are the reciprocals of the old ones.

Exam Tip: Using this coefficient swap trick can save you a lot of time on competitive exams.

Question 7. One of the roots of Q.E. \( x^2 - kx + 6 = 0 \) is 2. Find the other root.
Answer: Let the roots of the quadratic equation \( x^2 - kx + 6 = 0 \) be \( \alpha = 2 \) and \( \beta \). Comparing the equation with the standard form \( ax^2 + bx + c = 0 \), we have \( a = 1 \) and \( c = 6 \). The product of the roots of a quadratic equation is given by \( \frac{c}{a} \): \[ \alpha \times \beta = \frac{6}{1} \] Substituting the value of the given root \( \alpha = 2 \): \[ 2 \times \beta = 6 \] \[ \beta = 3 \] Thus, the other root of the quadratic equation is 3.
In simple words: Since the product of the roots is 6, and one root is 2, dividing 6 by 2 gives the other root, which is 3.

Exam Tip: Using the product of roots formula is much faster than solving for \( k \) first.

Question 8. One of the root of Q.E. is \( 2 + 3\sqrt{3} \), the other root is _________.
Answer: For a quadratic equation with rational coefficients, irrational roots always occur in conjugate pairs. This means that if one root is of the form \( a + b\sqrt{d} \), then the other root must be of the form \( a - b\sqrt{d} \). Given that one of the roots is \( 2 + 3\sqrt{3} \), the other root must be its conjugate, which is \( 2 - 3\sqrt{3} \).
In simple words: When a quadratic equation has regular fractions as coefficients, any root with a square root in it must have a matching partner with the opposite sign.

Exam Tip: Irrational roots always occur in conjugate pairs, so you only need to change the sign in front of the square root.

Question 9. One of the root of the Q.E. \( 3x^2 + 8x + k = 0 \) is the reciprocal of the other. Then k = _________.
Answer: Let the roots of the quadratic equation be \( \alpha \) and \( \frac{1}{\alpha} \). The product of the roots is given by the formula \( \frac{c}{a} \). For the given equation \( 3x^2 + 8x + k = 0 \), we have \( a = 3 \) and \( c = k \): \[ \text{Product of roots} = \alpha \times \frac{1}{\alpha} = \frac{k}{3} \] \[ 1 = \frac{k}{3} \implies k = 3 \] Therefore, the value of \( k \) is 3.
In simple words: Since the roots are reciprocals, their product is 1. This means the first and last coefficients must be equal, so \( k = 3 \).

Exam Tip: Whenever roots are reciprocals of each other, set the coefficient of \( x^2 \) equal to the constant term.

Question 10. \( \alpha \) and \( \beta \) are the roots of the Q.E. \( 5x^2 + 9x + 4 = 0 \), then the Q.E. whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is _________.
Answer: For the given quadratic equation \( 5x^2 + 9x + 4 = 0 \), we have \( a = 5 \), \( b = 9 \), and \( c = 4 \). We know that if a quadratic equation \( ax^2 + bx + c = 0 \) has roots \( \alpha \) and \( \beta \), then the equation with reciprocal roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is obtained by swapping the coefficients of \( x^2 \) and the constant term, resulting in \( cx^2 + bx + a = 0 \). Substituting our coefficients into this form: \[ 4x^2 + 9x + 5 = 0 \]
In simple words: To get reciprocal roots, simply swap the first and last numbers of the original equation to get \( 4x^2 + 9x + 5 = 0 \).

Exam Tip: Show the brief step-by-step verification using sum and product to ensure full credit.

Question 11. \( \alpha \) and \( \beta \) are the roots of the Q.E. \( x^2 - 5x + 6 = 0 \), then the Q.E. whose roots are \( 2\alpha \) and \( 2\beta \) is _________.
Answer: For the quadratic equation \( x^2 - 5x + 6 = 0 \): - The sum of roots \( \alpha + \beta = 5 \) - The product of roots \( \alpha\beta = 6 \) We want to find a new quadratic equation whose roots are \( 2\alpha \) and \( 2\beta \). Let's compute the sum and product for the new roots: - New Sum \( (S') = 2\alpha + 2\beta = 2(\alpha + \beta) = 2(5) = 10 \) - New Product \( (P') = (2\alpha)(2\beta) = 4\alpha\beta = 4(6) = 24 \) Using the standard formation of a quadratic equation \( x^2 - S'x + P' = 0 \): \[ x^2 - 10x + 24 = 0 \]
In simple words: For doubled roots, the sum of roots is doubled and the product of roots is quadrupled, which gives the equation \( x^2 - 10x + 24 = 0 \).

Exam Tip: When roots are multiplied by \( n \), the new equation is \( x^2 - nSx + n^2P = 0 \).

Question 12. The sides of a rectangle are x and x+2 cms and its area is \( 48 \text{ cm}^2 \). Find the perimeter of the rectangle. [Ans.: 28 cm]
Answer: Let the dimensions of the rectangle be represented by \( x \) and \( x + 2 \) centimeters. The area of a rectangle is calculated as the product of its length and width: \[ \text{Area} = x(x + 2) = 48 \] \[ x^2 + 2x - 48 = 0 \] We solve this quadratic equation by splitting the middle term: \[ x^2 + 8x - 6x - 48 = 0 \] \[ x(x + 8) - 6(x + 8) = 0 \] \[ (x - 6)(x + 8) = 0 \] Since side lengths must be positive, we reject \( x = -8 \). Thus, \( x = 6 \) cm. The dimensions of the rectangle are: - Width = \( 6 \) cm - Length = \( 6 + 2 = 8 \) cm Now, we calculate the perimeter of the rectangle: \[ \text{Perimeter} = 2(\text{Length} + \text{Width}) = 2(8 + 6) = 2(14) = 28 \text{ cm} \]
In simple words: Solve the equation \( x(x+2) = 48 \) to find the width is 6 cm. This makes the length 8 cm, and the perimeter is 28 cm.

Exam Tip: Always discard negative values for length or width since real-world measurements cannot be negative.

Question 13. The sum of 3 consecutive terms of an A.P is 24 and their product is 480. The terms are _________. [Ans.: 6, 8, 10]
Answer: Let the three consecutive terms of the Arithmetic Progression (A.P.) be \( a - d \), \( a \), and \( a + d \), where \( a \) is the middle term and \( d \) is the common difference. According to the first condition, the sum of these terms is 24: \[ (a - d) + a + (a + d) = 24 \] \[ 3a = 24 \implies a = 8 \] According to the second condition, the product of the terms is 480: \[ (a - d) \times a \times (a + d) = 480 \] Substituting the value of \( a = 8 \): \[ (8 - d) \times 8 \times (8 + d) = 480 \] \[ (8 - d)(8 + d) = \frac{480}{8} \] \[ 64 - d^2 = 60 \] \[ d^2 = 64 - 60 = 4 \] \[ d = \pm 2 \] - If \( d = 2 \), the terms are \( 8 - 2 = 6 \), \( 8 \), and \( 8 + 2 = 10 \). - If \( d = -2 \), the terms are \( 8 - (-2) = 10 \), \( 8 \), and \( 8 + (-2) = 6 \). In both scenarios, the three consecutive terms of the progression are 6, 8, and 10.
In simple words: Let the three terms be \( a-d \), \( a \), and \( a+d \). Their sum gives \( a = 8 \), and their product gives \( d = \pm 2 \), which gives the terms 6, 8, and 10.

Exam Tip: Using \( a-d, a, a+d \) simplifies the sum equation immensely by eliminating the common difference variable.

Question 14. The equation \( (k + 1)x^2 - 2(k - 1)x + 1 = 0 \) has equal roots. The value of k is _________. [Ans.: 0, 3]
Answer: For the given quadratic equation \( (k + 1)x^2 - 2(k - 1)x + 1 = 0 \) to have equal roots, its discriminant \( D \) must be equal to zero. Here, the coefficients are: - \( a = k + 1 \) - \( b = -2(k - 1) \) - \( c = 1 \) We set the discriminant \( D = b^2 - 4ac = 0 \): \[ [-2(k - 1)]^2 - 4(k + 1)(1) = 0 \] \[ 4(k - 1)^2 - 4(k + 1) = 0 \] Dividing by 4 on both sides: \[ (k - 1)^2 - (k + 1) = 0 \] \[ (k^2 - 2k + 1) - k - 1 = 0 \] \[ k^2 - 3k = 0 \] \[ k(k - 3) = 0 \] This gives \( k = 0 \) or \( k = 3 \). Both values of \( k \) satisfy the condition for the equation to have equal roots.
In simple words: Setting the discriminant of the equation to zero gives \( k^2 - 3k = 0 \), which gives \( k = 0 \) or \( k = 3 \).

Exam Tip: Factor out common numbers before solving the quadratic equation to simplify your calculations.

Question 15. The roots of the equation \( (b - c)x^2 + (c - a)x + (a - b) = 0 \) are equal. Prove that \( 2b = a + c \).
Answer: Let the given quadratic equation be \( (b - c)x^2 + (c - a)x + (a - b) = 0 \). First, let us observe the sum of the coefficients: \[ (b - c) + (c - a) + (a - b) = 0 \] Whenever the sum of the coefficients of a quadratic equation \( Ax^2 + Bx + C = 0 \) is zero, \( x = 1 \) is always one of the roots. Since the roots are given to be equal, both roots of this equation must be equal to 1. We know that the product of roots is given by the ratio of the constant term to the leading coefficient: \[ \text{Product of roots} = 1 \times 1 = \frac{a - b}{b - c} \] \[ 1 = \frac{a - b}{b - c} \] Cross-multiplying: \[ b - c = a - b \] Rearranging the terms: \[ b + b = a + c \] \[ 2b = a + c \] (Hence proved)
In simple words: Since the sum of the coefficients is zero, \( x = 1 \) is a root. Since the roots are equal, both must be 1, which leads to \( 2b = a+c \).

Exam Tip: This is a classic algebraic proof; recognizing that \( x = 1 \) is a root when coefficients sum to zero is a powerful shortcut.

Question 16. The sum of two nos. is 15 and the sum of their reciprocal is \( \frac{3}{10} \). Find the nos. [Ans.: 10, 5]
Answer: Let the two numbers be \( x \) and \( y \). According to the first condition: \[ x + y = 15 \implies y = 15 - x \] (Equation 1) According to the second condition: \[ \frac{1}{x} + \frac{1}{y} = \frac{3}{10} \] (Equation 2) Let's simplify Equation 2 by finding a common denominator: \[ \frac{x + y}{xy} = \frac{3}{10} \] Substitute \( x + y = 15 \) into this: \[ \frac{15}{xy} = \frac{3}{10} \] \[ 3xy = 150 \implies xy = 50 \] Now substitute \( y = 15 - x \) from Equation 1: \[ x(15 - x) = 50 \] \[ 15x - x^2 = 50 \] \[ x^2 - 15x + 50 = 0 \] Factorize the quadratic equation: \[ x^2 - 10x - 5x + 50 = 0 \] \[ x(x - 10) - 5(x - 10) = 0 \] \[ (x - 5)(x - 10) = 0 \] This gives \( x = 5 \) or \( x = 10 \). If \( x = 5 \), then \( y = 10 \); and if \( x = 10 \), then \( y = 5 \). Thus, the two required numbers are 5 and 10.
In simple words: Use the sum of reciprocals to find the product of the numbers is 50. Solving the system gives the two numbers as 5 and 10.

Exam Tip: Simplify the fractional sum of reciprocals first to find the product of the two numbers directly.

Question 17. The sum of the squares of two positive integers is 208. If the square if the larger number is 18 times the smaller no., find the nos.
Answer: Let the smaller positive integer be \( x \) and the larger positive integer be \( y \). According to the given conditions: 1. The sum of the squares of the two positive integers is 208: \[ x^2 + y^2 = 208 \] (Equation 1) 2. The square of the larger number is 18 times the smaller number: \[ y^2 = 18x \] (Equation 2) Substitute Equation 2 into Equation 1: \[ x^2 + 18x = 208 \] \[ x^2 + 18x - 208 = 0 \] We solve the quadratic equation by factoring: \[ x^2 + 26x - 8x - 208 = 0 \] \[ x(x + 26) - 8(x + 26) = 0 \] \[ (x - 8)(x + 26) = 0 \] Since \( x \) must be a positive integer, we reject the negative solution \( x = -26 \). Thus: \[ x = 8 \] Now, find \( y^2 \) using Equation 2: \[ y^2 = 18(8) = 144 \] Since \( y \) must be a positive integer: \[ y = \sqrt{144} = 12 \] Thus, the two positive integers are 8 and 12.
In simple words: Setting up the equations \( x^2 + y^2 = 208 \) and \( y^2 = 18x \) gives a quadratic equation that solves to \( x = 8 \) and \( y = 12 \).

Exam Tip: Be careful to discard negative integer solutions if the question specifies positive integers.

Question 18. The difference of the squares of two nos. is 45. If the square of the smaller no. is 4 times the larger no., find the nos.
Answer: Let the larger number be \( x \) and the smaller number be \( y \). According to the first condition, the difference of their squares is 45: \[ x^2 - y^2 = 45 \] (Equation 1) According to the second condition, the square of the smaller number is 4 times the larger number: \[ y^2 = 4x \] (Equation 2) Substitute Equation 2 into Equation 1: \[ x^2 - 4x = 45 \] \[ x^2 - 4x - 45 = 0 \] Factoring the quadratic equation: \[ x^2 - 9x + 5x - 45 = 0 \] \[ x(x - 9) + 5(x - 9) = 0 \] \[ (x - 9)(x + 5) = 0 \] This gives \( x = 9 \) or \( x = -5 \). - Case 1: If \( x = 9 \): \[ y^2 = 4(9) = 36 \implies y = \pm 6 \] This yields the number pairs: \( (9, 6) \) and \( (9, -6) \). - Case 2: If \( x = -5 \): \[ y^2 = 4(-5) = -20 \] Since the square of a real number cannot be negative, this case has no real solution. Therefore, the numbers are either 9 and 6, or 9 and -6.
In simple words: Setting up the equations \( x^2 - y^2 = 45 \) and \( y^2 = 4x \) gives the numbers as 9 and 6, or 9 and -6.

Exam Tip: Check if negative values are allowed; here, "two numbers" allows negative values, so \( \pm 6 \) are both valid.

Question 19. The hypotenuse of a right angled triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse. Find the sides.
Answer: Let the length of the shortest side of the right-angled triangle be \( x \) meters. Based on the given conditions, we can express the other two sides as follows: - Hypotenuse = \( 2x + 6 \) meters - Third side = \( (2x + 6) - 2 = 2x + 4 \) meters By applying the Pythagoras theorem: \[ (\text{Shortest side})^2 + (\text{Third side})^2 = (\text{Hypotenuse})^2 \] \[ x^2 + (2x + 4)^2 = (2x + 6)^2 \] Expanding both squared binomials: \[ x^2 + (4x^2 + 16x + 16) = 4x^2 + 24x + 36 \] Combine like terms: \[ 5x^2 + 16x + 16 = 4x^2 + 24x + 36 \] Rearranging into a standard quadratic equation: \[ x^2 - 8x - 20 = 0 \] Factoring by splitting the middle term: \[ x^2 - 10x + 2x - 20 = 0 \] \[ x(x - 10) + 2(x - 10) = 0 \] \[ (x - 10)(x + 2) = 0 \] Since side lengths must be positive, we reject \( x = -2 \). Thus, \( x = 10 \). Now, find the lengths of all three sides: - Shortest side = \( 10 \) m - Third side = \( 2(10) + 4 = 24 \) m - Hypotenuse = \( 2(10) + 6 = 26 \) m The side lengths of the triangle are 10 m, 24 m, and 26 m.
In simple words: Using Pythagoras theorem on sides \( x \), \( 2x+4 \), and \( 2x+6 \) gives \( x = 10 \), making the sides 10 m, 24 m, and 26 m.

Exam Tip: Be careful to expand binomial squares like \( (2x+4)^2 \) using the full identity \( a^2 + 2ab + b^2 \).

Question 20. The perimeter of a rectangle is 70 cm and its diagonal is 25 cm. Find the area of the rectangle.
Answer: Let the length and width of the rectangle be \( l \) and \( w \) respectively. The perimeter of the rectangle is given as 70 cm: \[ 2(l + w) = 70 \implies l + w = 35 \implies w = 35 - l \] (Equation 1) The length of the diagonal is given as 25 cm. Using Pythagoras theorem: \[ l^2 + w^2 = 25^2 = 625 \] (Equation 2) Substitute Equation 1 into Equation 2: \[ l^2 + (35 - l)^2 = 625 \] \[ l^2 + (1225 - 70l + l^2) = 625 \] \[ 2l^2 - 70l + 1225 = 625 \] \[ 2l^2 - 70l + 600 = 0 \] Dividing by 2 on both sides: \[ l^2 - 35l + 300 = 0 \] Factoring the quadratic equation: \[ l^2 - 20l - 15l + 300 = 0 \] \[ l(l - 20) - 15(l - 20) = 0 \] \[ (l - 15)(l - 20) = 0 \] This gives \( l = 15 \) cm or \( l = 20 \) cm. - If length \( l = 20 \) cm, then width \( w = 15 \) cm. - If length \( l = 15 \) cm, then width \( w = 20 \) cm. The area of the rectangle is: \[ \text{Area} = l \times w = 20 \times 15 = 300 \text{ cm}^2 \]
In simple words: The sum of length and width is 35 cm. Using the diagonal of 25 cm with Pythagoras theorem gives the area as 300 square centimeters.

Exam Tip: You can also solve this using the identity \( (l+w)^2 = l^2 + w^2 + 2lw \) without solving for individual sides.

Question 21. Two pipes running together can fill a cistern in \( 3 \frac{1}{13} \) minutes. If one of the pipes takes 3 minutes less than the other to fill the cistern, find the time taken by each of them to fill the tank.
Answer: Let the time taken by the slower pipe to fill the cistern alone be \( x \) minutes. Then, the time taken by the faster pipe is \( x - 3 \) minutes. In one minute: - Slower pipe fills \( \frac{1}{x} \) of the cistern. - Faster pipe fills \( \frac{1}{x - 3} \) of the cistern. Together, the two pipes fill the cistern in \( 3 \frac{1}{13} = \frac{40}{13} \) minutes, which means in one minute they fill \( \frac{13}{40} \) of the cistern. Thus, we set up the equation: \[ \frac{1}{x} + \frac{1}{x - 3} = \frac{13}{40} \] \[ \frac{(x - 3) + x}{x(x - 3)} = \frac{13}{40} \] \[ \frac{2x - 3}{x^2 - 3x} = \frac{13}{40} \] Cross-multiplying: \[ 40(2x - 3) = 13(x^2 - 3x) \] \[ 80x - 120 = 13x^2 - 39x \] \[ 13x^2 - 119x + 120 = 0 \] We solve this quadratic equation by splitting the middle term: \[ 13x^2 - 104x - 15x + 120 = 0 \] \[ 13x(x - 8) - 15(x - 8) = 0 \] \[ (13x - 15)(x - 8) = 0 \] This gives \( x = 8 \) or \( x = \frac{15}{13} \). Since the faster pipe takes \( x - 3 \) minutes, if \( x = \frac{15}{13} \approx 1.15 \), the time taken would be negative, which is impossible. Therefore, we choose \( x = 8 \) minutes. - Slower pipe's time = 8 minutes - Faster pipe's time = \( 8 - 3 = 5 \) minutes So, the slower pipe takes 8 minutes and the faster pipe takes 5 minutes to fill the cistern individually.
In simple words: Setting up the rate of work equation \( \frac{1}{x} + \frac{1}{x-3} = \frac{13}{40} \) yields \( x = 8 \), so the times are 8 minutes and 5 minutes.

Exam Tip: Always check if the fractional solution of \( x \) makes any of the side times negative, and discard it if so.

Question 22. A dealer sells a toy for Rs. 24 and gains as much as percent as the cost price (in Rs.). Find the Cost Price.
Answer: Let the cost price (C.P.) of the toy be Rs. \( x \). The profit percentage is equal to the numerical value of the cost price, which is \( x\% \). Therefore, the profit amount is: \[ \text{Profit} = x\% \text{ of C.P.} = \frac{x}{100} \times x = \frac{x^2}{100} \] The selling price (S.P.) is given by the sum of cost price and profit: \[ \text{S.P.} = \text{C.P.} + \text{Profit} \] \[ 24 = x + \frac{x^2}{100} \] Multiply the entire equation by 100 to eliminate the denominator: \[ 2400 = 100x + x^2 \] \[ x^2 + 100x - 2400 = 0 \] Factoring the quadratic equation: \[ x^2 + 120x - 20x - 2400 = 0 \] \[ x(x + 120) - 20(x + 120) = 0 \] \[ (x - 20)(x + 120) = 0 \] Since cost price cannot be negative, we reject \( x = -120 \). Thus, \( x = 20 \). The Cost Price of the toy is Rs. 20.
In simple words: Set up the equation \( x + \frac{x^2}{100} = 24 \). Solving this quadratic equation gives the cost price of the toy as Rs. 20.

Exam Tip: In profit percentage problems, remember to express the percentage profit relative to the cost price as \( \frac{x^2}{100} \).

Question 23. A librarian bought some story books for children for Rs. 2000. Had the price of each book been reduced by Rs. 10, he would have bought 10 more books. Find the ORIGINAL price of each book.
Answer: Let the original price of each story book be Rs. \( x \). The number of books the librarian can buy with Rs. 2000 originally is: \[ N_1 = \frac{2000}{x} \] If the price of each book is reduced by Rs. 10, the new price is Rs. \( x - 10 \). The number of books that can be bought with Rs. 2000 at this reduced price is: \[ N_2 = \frac{2000}{x - 10} \] According to the problem, the difference between the two quantities is 10 books: \[ N_2 - N_1 = 10 \] \[ \frac{2000}{x - 10} - \frac{2000}{x} = 10 \] Dividing by 10 on both sides: \[ \frac{200}{x - 10} - \frac{200}{x} = 1 \] Finding a common denominator: \[ 200 \left( \frac{x - (x - 10)}{x(x - 10)} \right) = 1 \] \[ 200 \left( \frac{10}{x^2 - 10x} \right) = 1 \] \[ 2000 = x^2 - 10x \] \[ x^2 - 10x - 2000 = 0 \] Factoring the quadratic equation: \[ x^2 - 50x + 40x - 2000 = 0 \] \[ x(x - 50) + 40(x - 50) = 0 \] \[ (x - 50)(x + 40) = 0 \] Since the price of a book cannot be negative, we discard \( x = -40 \). Thus: \[ x = 50 \] The original price of each story book is Rs. 50.
In simple words: Setting up the book equation \( \frac{2000}{x-10} - \frac{2000}{x} = 10 \) gives the original price of each story book as Rs. 50.

Exam Tip: Reduce both sides of the fractional equation by a common divisor (like 10) early on to keep the numbers manageable.