CBSE Class 10 Mathematics Quadratic Equation Worksheet Set P

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.