CBSE Class 10 Mathematics Quadratic Equations Worksheet Set B

VERY SHORT ANSWER TYPE QUESTIONS :

Question. If one of the roots of x² + px − 4 = 0 is − 4 then find the product of its roots and the value of p.
Answer : If − 4 is a root of the quadratic equation,
x² + px − 4 = 0
∴ (− 4)² + (− 4) (p) − 4 = 0
⇒ 16 − 4p − 4 = 0
⇒ 12 − 4p = 0 ⇒ p = 3
Now, in ax² + bx + c = 0, the product of the roots = c/a
∴ Product of the roots in x² − px − 4 = 0
= − 4/1 = − 4

Question. For what value of k are the roots of the quadratic equation kx² + 4x + 1 = 0 equal and real?
Answer : Comparing kx² + 4x + 1 = 0, with ax² + bx + c = 0, we get
a = k
b = 4
c = 1
∴ b² − 4ac = (4)² − 4 (k) (1)
= 16 − 4k
For equal and real roots, we have
b² − 4ac = 0
⇒ 16 − 4k = 0
⇒ 4k = 16
⇒ k = 16/4 = 4

Question. Using quadratic formula, solve the following quadratic equation for x:
x² − 2ax + (a² − b²) = 0
Answer : Comparing x² − 2ax + (a² − b²) = 0, with ax² + bx + c = 0, we have:
a = 1, b = − 2a, c = a² − b²
∴ x = − b ± √b² − 4ac/2a
⇒ x = − (− 2a) ± − √(2a)² − 4 (1) (a² − b²)/2)1_
= 2a ± √4a² − 4a² − 4b²
= 2a ± √4b²/2 = 2a ± b/2 = a ± b
∴ x = (a + b) or x = (a − b)

Question. For what value of k, does the quadratic equation x² − kx + 4 = 0 have equal roots?
Answer : Comparing x² − kx + 4 = 0 with ax² + bx + c = 0, we get
a = 1
b = − k
c = 4
∴ b² − 4ac = (− k)² − 4 (1) (4) = k² − 16
For equal roots,
b² − 4ac = 0
⇒ k² − 16 = 0
⇒ k² = 16
⇒ k = ± 4

Question. If the roots of the quadratic equation − ax² + bx + c = 0 are equal then show that b² = 4ac.
Answer : ä For equal roots, we have
b² − 4ac = 0
∴ b²= 4ac

Question. If one root of the quadratic equation 2x² − 3x + p = 0 is 3, find the other root of the quadratic equation. Also find the value of p.
Answer : We have:
2x²− 3x + p = 0
∴ a = 2, b = − 3 and c = p
Since, the sum of the roots = − b/a
= − (−3)/2 = 3/2
∴ One of the roots = 3
∴ The other root = 3/2 − 3 = − 3/2
Now, substituting x = 3 in (1), we get
2 (3)² − 3 (3) + p = 0
⇒ 18 − 9 + p = 0
⇒ 9 + p = 0 ⇒ p = − 9

Question. For what value of k does (k − 12) x² + 2 (k − 12) x + 2 = 0 have equal roots?
Answer : Comparing (k − 12) x² + 2 (k − 12) x + 2 = 0 with ax² + bx + c = 0, we have:
a = (k − 12)
b = 2 (k − 12)
c = 2
∴ b² − 4ac = [2 (k − 12)]² − 4 (k − 12) (2)
= 4 (k − 12)² − 8 (k − 12)
= 4 (k − 12) [k − 12 − 2]
= 4 (k − 12) (k − 14)
For equal roots,
b² − 4ac = 0
⇒ 4 (k − 12) [k − 14] = 0
⇒ Either 4 (k − 12) = 0 ⇒ k = 12
or k − 14 = 0 ⇒ k = 14
But k = 12 makes k − 12 = 0 which is not required
∴ k ≠ 12
⇒ k = 14

Question. For what value of k, does the given equation have real and equal roots?
(k + 1) x² − 2 (k − 1) x + 1 = 0.
Answer : Comparing the given equation with ax² + bx + c = 0, we have:
a = k + 1
b = − 2 (k − 1)
c = 1
For equal roots, b² − 4ac = 0
∴ [− 2 (k − 1)]² − 4 (k + 1) (1) = 0
⇒ 4 (k − 1)² − 4 (k + 1) = 0
⇒ 4 (k² + 1 − 2k) − 4k − 4 = 0
⇒ 4k² + 4 − 8k − 4k − 4 = 0
⇒ 4k² − 12k = 0
⇒ 4k (k − 3) = 0
⇒ k = 0 or k = 3

Question. For what value of k does the equation 9x² + 3kx + 4 = 0 has equal roots?
Answer : Comparing 9x² + 3kx + 4 = 0 with ax² + bx + c = 0, we get
a = 9
b = 3k
c = 4
∴ b² − 4ac = (3k)² − 4 (9) (4)
= 9k² − 144
For equal roots,
b² − 4ac = 0
⇒ 9k² − 144 = 0
⇒ 9k² = 144
⇒ k² = 144/9 = 16
⇒ k = ± 4

Question. What is the nature of roots of the quadratic equation 4x² − 12x + 9 = 0?
Answer : Comparing 4x² − 12x + 9 = 0 with ax² + bx + c = 0 we get
a = 4
b = − 12
c = 9
∴ b² − 4ac = (− 12)² − 4 (4) (9)
= 144 − 144 = 0
Since b² − 4ac = 0
∴ The roots are real and equal.

Question. If one of the roots of the quadratic equation 2x² + kx − 6 = 0 is 2, find the value of k. Also find the other root.
Answer : Given equation:
2x² + kx − 6 = 0
one root = 2
Substituting x = 2 in 2x² + kx − 6 = 0
We have:
2 (2)² + k (2) − 6 = 0
⇒ 8 + 2k − 6 = 0
⇒ 2k + 2 = 0 ⇒ k = − 1
∴ 2x² + kx − 6 = 0 ⇒ 2x² − x − 6 = 0
Sum of the roots = − b/a = 1/2
∴ other root = 1/2 − 2
= 3/2

Question. Find the value of ‘k’ for which the quadratic equation kx² − 5x + k = 0 have real roots.
Answer : Comparing kx² − 5x + k = 0 with ax²+ bx + c = 0, we have:
a = k
b = − 5
c = k
∴ b² − 4ac = (− 5)² − 4 (k) (k)
= 25 − 4k²
For equal roots, b² − 4ac = 0
∴ 25 − 4k² = 0
⇒ 4k² = 25
⇒ k² = 25/4
⇒ k = ±√25/4 = ±5/2

Question. Write the value of k for which the quadratic equation x² − kx + 9 = 0 has equal roots.
Answer : Comparing x² − kx + 9 = 0 with ax² + bx + c = 0, we get
a = 1
b = − k
c = 9
∴ b² − 4ac = (− k)² − 4 (1) (9)
= k² − 36
For equal roots,
b² − 4ac = 0
⇒ k² − 36 = 0 ⇒ k² = 36
⇒ k = ± 6

Question. If − 4 is a root of the quadratic equation x² + px − 4 = 0 and x² + px + k = 0 has equal roots, find the value of k.
Answer : ∴ (–4) is a root of x² + px − 4 = 0
∴ (− 4)² + p (− 4) = 0
⇒ 16 − 4p − 4 = 0
⇒ 4p = 12 or p = 3
Now, x²+ px + k = 0
⇒ x² + 3x + k = 0 [ ∴ p = 3]
Now, a = 1, b = 3 and c = + k
∴ b² − 4ac = (3)2 − 4 (1) (k)
= 9 − 4k
For equal roots, b² − 4ac = 0
⇒ 9 − 4k = 0 ⇒ 4k = 9
⇒ k = 9/4

Question. Determine the value of k for which the quadratic equation 4x² − 4kx + 1 = 0 has equal roots.
Answer : We have:
4x² − 4kx + 1 = 0
Comparing with ax² + bx + c = 0, we have
a = 4, b = − 4k and c = 1
∴ b² − 4ac = (− 4k)² − 4 (4k) (1)
= 16k² − 16
For equal roots
b² − 4ac = 0
∴ 16k² − 16 = 0
⇒ 16k² = 16 ⇒ k² = 1
⇒ k = ± 1

Question. For what value of k are the roots of the quadratic equation 3x² + 2kx + 27 = 0 real and equal?
Answer : Comparing 3x² + 2 kx + 27 = 0 with ax²+ bx + c = 0, we have:
a = 3
b = 2k
c = 27
∴ b² − 4ac = (2k)² − 4 (3) (27)
= 4k² − (12 × 27)
For the roots to be real and equal
b² − 4ac = 0
⇒ 4k² − (12 × 17) = 0
⇒ 4k² = 12 × 27
⇒ k² = 12 × 27/4 = 81
⇒ k = ± 9

SHORT ANSWER TYPE QUESTIONS :

Question. Solve for x: 36x² − 12ax + (a² − b²) = 0.
Answer : We have:
36x² − 12ax + (a² − b²) = 0 
Comparing (1) with Ax² + Bx + C = 0, we have:
A = 36
B = − 12a
C = (a2 − b2)
∴ B² − 4AC = [− 12a]² − 4 (36) [a² − b²]
= 144 a² − 144 (a² − b²)
= 144 a² − 144 a² + 144 b²
= 144 b²
Since, x = − B ± √B² − 4AC/2A
∴ x = − (− 12a) ± √144 b²/2 (36)
⇒ x = 12a ± 12b/72
Taking +ve sign, we have:
x = 12a + 12b/72
⇒ x = 12/72 (a + b) = 1/6 (a + b)
Taking −ve sign, we get
x = 12a − 12b/72 (a − b) = 1/6 (a − b)
Thus, the required roots are:
x = 1/6 (a + b) and x = 1/6 (a − b)

Question. Find the roots of the quadratic equation 2x² - √5x - 2 = 0, using the quadratic formula.
Answer : 

Question. Find the roots of the equation:
1/x − 2 + 1/x = 8/2x + 5 : x ≠ 0, 2 −5/2
Answer : We have:

Question. Solve 2x² − 5x + 3 = 0.
Answer : We have:
2x² − 5x + 3 = 0 
Comparing (1) with ax2 + bx + c = 0,
∴ a = 2
b = − 5
c = 3
∴ b² − 4ac = (− 5)² − 4 (2) (3)
= 25 − 24 = 1
Since, x = − b ± √b² − 4ac/2a
∴ x = − (− 5) ± √1/2 (2)
= 5 ± 1/4
Taking, + ve sign,
x = 5 + 1/4 = 6/4 = 3/2
Taking, −ve sign,
x = 5 − 1/4 = 4/4 = 1
Thus, the required roots are
x = 3/2 and x = 1

Question. Solve for x: 9x² − 6ax + a² − b² = 0.
Answer : We have:
9x² − 6ax + (a² − b²) = 0 
Comparing (1) with ax² + bx + c = 0, we get
a = 9, b = − 6a and c = (a2 − b2)
∴ b² − 4ac = (− 6a)² − 4 (9) (a² − b²)
= 36a²  − 36 (a²  − b²)
= 36a²  − 36a²  + 36b²
= 36b² = (6b)²
Since, x = − b ± √b² − 4ac/2a
∴ x = − (− 6a) ± 6√(6b)²/2(9)
⇒ x = 6a ± 6b/18
⇒ x = 6[a ± b]/18 = a ± b/3
a ± b a b
Taking the +ve sign, we get
x = a + b/3
Taking the −ve sign, we get
x = a − b/3
∴ The required roots are:
x = a + b/3 and x = a − b/3

Question. Evaluate √20 + √20 + √20 + 
Answer : 

Question. Find the roots of the equation:
1/x − 1/x − 3 = 4/3  ; x ≠ 0, 3
Answer : We have:

Question. Solve the following quadratic equation:
2x² + 4x − 8 = 0
Answer : We have:
2x² + 4x − 8 = 0
Dividing by 2, we get
x²+ 2x − 4 = 0 
Comparing (1) with ax2 + bx + c = 0,
a = 1,
b = 2
c = − 4
∴ b² − 4ac = (2)² − 4 (1) (− 4)
= 4 + 16 = 20
Since, x = − b ± √b² − 4ac/2a
∴ x =
− 2 ± √20/2 (1) 
⇒ x = − 2 ± 2√5/2
⇒ x = 2 [−1 ± √5]/2  = − 1 ± √5
Taking + ve sign, we get
x = [− 1 + √5]
Taking − ve sign, we get
x = [− 1 − √5]
Thus, the required roots are x = [− 1 + √5] and x = [− 1 − √5]

Question. Using quadratic formula, solve the following quadratic equation for x:
x² − 4ax + 4a² − b² = 0
Answer : Comparing the given equation with ax² + bx + c = 0, we have:
a = 1
b = − 4a
c = 4a² − b²
∴ b² − 4ac = [− (4a)]² − 4 (1) [4a² − b²]
= 16a² − 4 (4a² − b²)
= 16a² − 16a² + 4b²
= 4b² = (2b)²
Since, x = − b ± √b² − 4ac/2a
∴ x = − (− 4a) ± √(2b)²/2 (1)
⇒ x = 4a ± 2b/2
⇒ x = 2/2 [2a ± b] = 2a ± b
Taking the +ve sign, x = 2a + b
Taking the −ve sign, x = 2a − b
Thus, the required roots are:
x = 2a + b and x = 2a − b

Question. Using quadratic formula, solve the following quadratic equation for x:
x² − 2ax + (a² − b²) = 0
Answer : Comparing the given equation with ax² + bx + c = 0, we have:

Question. Solve: 16x² − 8a² x + (a⁴ − b⁴) = 0 for x.
Answer : We have:
16x² − 8a² x + a⁴ − b⁴ = 0 
Comparing (1) with ax² + bx + c = 0, we get
a = 16
b = − 8a²
c = (a⁴ − b⁴)
∴ b² − 4ac = [− 8a²]² − 4 (16) (a⁴ − b⁴)
= 64 a⁴ − 64 (a⁴ − b⁴)
= 64 a⁴ − 64 a⁴ + 64 b⁴
= 64 b⁴

Question. Solve (using quadratic formula):
x² + 5x + 5 = 0
Answer : We have:
x² + 5x + 5 = 0
Comparing (1) with ax² + bx + c = 0, we have:
a = 1
b = 5
c = 5
∴ b² − 4ac = (5)² − 4 (1) (5)
= 25 − 20 = 5
Since, x = − b ± √b² − 4ac/2a
∴ x = − 5 ± √5/2 (1)
⇒ x = − 5 ± √5/2
Taking +ve sign, we have:
x = − 5 + √5/2
Taking −ve sign, we have:
x = − 5 − √5/2
Thus, the required roots are:
x = − 5 + √5/2 and x = − 5 − √5/2

Question. Find the roots of the following equation:
1/x + 4 − 1/x − 7 = 11/30; x = 4, 7
Answer : We have:
1/x + 4 − 1/x − 7 = 11/30

Extra Based Questions :

Check whether the following are quadratic equations:

Question. (x + 1)² = 2(x − 3)
Answer : (x + 1)² = 2(x − 3)
We have:
(x + 1)2 = 2 (x − 3)
⇒ x² + 2x + 1 = 2x − 6
⇒ x² + 2x + 1 − 2x + 6 = 0
⇒ x² + 7 = 0
Since x² + 7 is a quadratic polynomial
∴ (x + 1)² = 2(x − 3) is a quadratic equation.

Question. x² − 2x = (−2) (3 − x)
Answer : x² − 2x = (− 2) (3 − x)
We have:
x² − 2x = (− 2) (3 − x)
⇒ x² − 2x = − 6 + 2x
⇒ x² − 2x − 2x + 6 = 0
⇒ x² − 4x + 6 = 0
Since x² − 4x + 6 is a quadratic polynomial
∴ x² − 2x = (−2) (3 − x) is a quadratic equation.

Question. (x − 2) (x + 1) = (x − 1) (x + 3)
Answer : (x − 2) (x + 1) = (x − 1) (x + 3)
We have:
(x − 2) (x + 1) = (x − 1) (x + 3)
⇒ x² − x − 2 = x² + 2x − 3
⇒ x² − x − 2 − x² − 2x + 3 = 0
⇒ −3x + 1 = 0
Since −3x + 1 is a linear polynomial
∴ (x − 2) (x + 1) = (x − 1) (x + 3) is not quadratic equation.

Question. (x − 3) (2x + 1) = x(x + 5)
Answer : (x − 3) (2x + 1) = x(x + 5)
We have:
(x − 3) (2x + 1) = x (x + 5)
⇒ 2x² + x − 6x − 3 = x² + 5x
⇒ 2x² − 5x − 3 − x² − 5x = 0
⇒ x² + 10x − 3 = 0
Since x² + 10x − 3 is a quadratic polynomial
∴ (x − 3) (2x + 1) = x(x + 5) is a quadratic equation.

Question. (2x − 1) (x − 3) = (x + 5) (x − 1)
Answer : (2x − 1) (x − 3) = (x + 5) (x − 1)
We have:
(2x − 1) (x − 3) = (x + 5) (x − 1)
⇒ 2x² − 6x − x + 3 = x² − x + 5x − 5
⇒ 2x² − x² − 6x − x + x − 5x + 3 + 5 = 0
⇒ x² − 11x + 8 = 0
Since x² − 11x + 8 is a quadratic polynomial
∴ (2x − 1) (x − 3) = (x + 5) (x − 1) is a quadratic equation.

Question. x² + 3x + 1 = (x − 2)²
Answer : x² + 3x + 1 = (x − 2)²
We have:
x²+ 3x + 1 = (x − 2)²
⇒ x² + 3x + 1 = x² − 4x + 4
⇒ x² + 3x + 1 − x² + 4x − 4 = 0
⇒ 7x − 3 = 0
Since 7x − 3 is a linear polynomial.
∴ x² + 3x + 1 = (x − 2)² is not a quadratic equation.

Question. (x + 2)³ = 2x(x² − 1)
Answer : (x + 2)³ = 2x(x²− 1)
We have:
(x + 2)³ = 2x(x² − 1)
⇒ x³ + 3x²(2) + 3x(2)² + (2)³ = 2x³ − 2x
⇒ x³ + 6x² + 12x + 8 = 2x³ − 2x
⇒ x³ + 6x² + 12x + 8 − 2x³ + 2x = 0
⇒ − x³ + 6x² + 14x + 8 = 0
Since −x³ + 6x² + 14x + 8 is a polynomial of degree 3
∴ (x + 2)³ = 2x(x² − 1) is not a quadratic equation.

Question. x³ − 4x² − x + 1 = (x − 2)³
Answer : x³ − 4x² − x + 1 = (x − 2)³
We have:
x³ − 4x² − x + 1 = (x − 2)³
⇒ x³ − 4x² − x + 1 = x³ + 3x²(− 2) + 3x(− 2)² + (− 2)³
⇒ x³ − 4x² − x + 1 = x³ − 6x² + 12x − 8
⇒ x³− 4x² − x − 1 − x³ + 6x² − 12x + 8 = 0
⇒ 2x² − 13x + 9 = 0
Since 2x² − 13x + 9 is a quadratic polynomial
∴ x³ − 4x² − x + 1 = (x − 2)³ is a quadratic equation.

Represent the following situations in the form of quadratic equations:

Question. The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Answer : Let the breadth = x metres
ä Length = 2 (Breadth) + 1
∴ Length = (2x + 1) metres
Since Length × Breadth = Area
∴ (2x + 1) × x = 528
⇒ 2x² + x = 528
⇒ 2x² + x − 528 = 0
Thus, the required quadratic equation is
2x² + x − 528 = 0

Question. The product of two consecutive positive integers is 306. We need to find the integers.
Answer : Let the two consecutive numbers be x and (x + 1).
QProduct of the numbers = 306
∴ x (x + 1) = 306
⇒ x² + x = 306
⇒ x² + x − 306 = 0
Thus, the required equdratic equation is
x² + x − 306 = 0

Question. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Answer : Let the present age = x
∴ Mother’s age = (x + 26) years
After 3 years
His age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
= (x + 29) years
According to the condition,
Product of their ages after 3 years    = 360
⇒ (x + 3) × (x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 29x + 3x + 87 − 360 = 0
⇒ x² + 32x − 273 = 0
Thus, the required quadratic equation is
x² + 32x − 273 = 0

Question. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer : Let the speed of the train = u km/hr
Distance covered = 480 km
Time taken = Distance ÷ Speed
= (480 ÷ u) hours
= 480/u hours
In second case,
Speed = (u − 8) km/hour
∴ Time taken = Distance/speed = 480/u − 8 hours
According to the condition,
480/u − 8 u − 8 480/u hour
⇒ 480u − 480(u − 8) = 3u(u − 8)
⇒ 480u − 480u + 3840 = 3u² − 24u
⇒ 3840 − 3u² + 24u = 0
⇒ 1280 − u² + 8u = 0
⇒ − 1280 + u² − 8u = 0
⇒ u² − 8u − 1280 = 0
Thus, the required quadratic equation is
u² − 8u – 1280 = 0

Find the roots of the following quadratic equations by factorisation:

Question. x² − 3x − 10 = 0
Answer : x² − 3x − 10 = 0
We have:
x² − 3x − 10 = 0
⇒ x² − 5x + 2x − 10 = 0
⇒ x (x − 5) + 2 (x − 5) = 0
⇒ (x − 5) (x + 2) = 0
Either x − 5 = 0 ⇒ x = 5
or x + 2 = 0 ⇒ x = − 2
Thus, the required roots are x = 5 and x = −2.

Question. 2x² + x − 6 = 0
Answer : 2x² + x − 6 = 0
We have:
2x² + x − 6 = 0
⇒ 2x² + 4x − 3x − 6 = 0
⇒ 2x(x + 2) − 3 (x + 2) = 0
⇒ (x + 2) (2x − 3) = 0
Either x + 2 = 0 ⇒ x = −2
or 2x − 3 = 0 ⇒ x = 3/2
Thus, the required roots are x = −2 and x = 3/2 .

Question. √2x² + 7x + 5√2 = 0
Answer : √2x² + 7x + 5√2 = 0
We have:
√2x² + 7x + 5√2 = 0
⇒ √2x² + 2x + 5x + 5√2 = 0
⇒ √2x² + (√2⋅√2) x + 5x + 5.√2 = 0
⇒ √2x [x + √2] + 5 [x + √2] = 0
⇒ (x + √2) (√2 + 5) = 0
Either x + √2 = 0 ⇒ x = −√2
or √2x + 5 = 0 ⇒ x = − 5/√2
Thus, the required roots are x = −√2 and x = − 5/√2.

Question. 2x² −x + 1/8 = 0
Answer : 2x² − x +x²
We have:
2x² − x + x²
⇒ 16x² − 8x + 1 = 0
⇒ 16x² − 4x − 4x + 1 = 0
⇒ 4x(4x − 1) − 1(4x − 1) = 0
⇒ (4x − 1) (4x − 1) = 0
⇒ x = 1/4 and x = 1/4
Thus, the required roots are x = 1/4 and x = 1/4.

Question. 100x² − 20x + 1 = 0
Answer : 100x² − 20x + 1 = 0
We have:
100x² − 20x + 1 = 0
⇒ 100x² − 10x − 10x + 1 = 0
⇒ 10x (10x − 1) − 1 (10x − 1) = 0
⇒ (10x − 1) (10x − 1) = 0
⇒ (10x − 1) = 0 and (10x − 1) = 0
⇒ x = 1/10 and x = 1/10 
Thus, the required roots are x = 1/10 and x = 1/10.

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

Question. 2x² − 3x + 5 = 0
Answer : 2x² − 3x + 5 = 0
Comparing the given quadratic equation with ax² + bx + c = 0, we have:
a = 2
b = − 3
c = 5
∴ The discriminant = b² − 4ac
= (− 3)² − 4 (2) (5)
= 9 − 40
= − 31 < 0
Since b² − 4ac is negative.
∴ The given quadratic equation has no real roots.

Question. 3x² − 4√3 x + 4 = 0
Answer : 3x² − 4√3 x + 4 = 0
Comparing the given quadratic equation with ax² + bx + c = 0, we get
a = 3
b = −4√3
c = 4
∴ b² − 4ac = [− 4√3] − 4(3)(4)
2 a fa f
= (16 × 3) − 48
= 48 − 48 = 0
Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:
− b/2a and − b/2a
i.e., −(−4√3)/2 × 3 and −(−4√3)/2 × 3
⇒ 4√3/2√3 × √3 and 4√3/2√3 × √3
⇒ 2/√3 and 2/√3
Thus, x = 2/√3 and x = 2/√3

Question. 2x² − 6x + 3 = 0
Answer : 2x² − 6x + 3 = 0
Comparing it with the general quadratic equation, we have:
a = 2
b = − 6
c = 3
∴ b² − 4ac = (− 6)² − 4 (2) (3)
= 36 − 24
= 12 > 0
∴ The given quadratic equation has two real and distinct roots, which are given by
x = − b ± √b² − 4ac/2a
⇒ x = − (−6) ± √12/2 × 2
= 6 ± 2 √3/4
= 3 ± √3/2
Thus, the roots are:
x = 3 + √3/2 and x = 3 − √3/2

Find the values of k for each of the following quadratic equations, so that they have two equal roots:

Question. 2x² + kx + 3 = 0
Answer : 2x² + kx + 3 = 0
Comparing the given quadratic equation with ax² + bx + c = 0, we get
a = 2
b = k
c = 3
∴ b² − 4ac = (− k)² − 4 (2) (3)
= k² − 24
Q For a quadratic equation to have equal roots,
b² − 4ac = 0
∴ k² − 24 = 0 ⇒ k = ± √24
⇒ k = ±2√6
Thus, the required values of k are
2√6 and −2√6

Question. kx (x − 2) + 6 = 0
Answer : kx (x − 2) + 6 = 0
Comparing kx (x − 2) + 6 = 0 i.e., kx² − 2kx + 6 = 0 with ax² + bx + c = 0, we get
a = k
b = − 2k
c = 6
∴ b² − 4ac = (− 2k)² − 4 (k) (6)
= 4k² − 24k
Since, the roots are real and equal,
∴ b² − 4ac = 0
⇒ 4k² − 24k = 0
⇒ 4k (k − 6) = 0
⇒ 4k = 0 or k − 6 = 0
⇒ k = 0 or k = 6
But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.

Question. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m? If so, find its length and breadth.
Answer : Let the breadth be x metres.
∴ Length = 2x metres
Now, Area = Length × Breadth
= 2x × x metre²
= 2x² sq. metre.
According to the given condition,
2x² = 800
⇒ x² = 800/2 = 400
⇒ x = ± √400 = ± 20
Therefore, x = 20 and x = − 20
But x = − 20 is possible                                                        (∴ breadth cannot be negative).
∴ x = 20
⇒ 2x = 2 × 20 = 40
Thus, length = 40 m and breadth = 20 m

Question. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer : Let the age of one friend = x years
∴ The age of the other friend = (20 − x) years                              [∴ Sum of their ages is 20 years]
Four years ago
Age of one friend = (x − 4) years
Age of the other friend = (20 − x − 4) years
= (16 − x) years
According to the condition,
(x − 4) × (16 − x) = 48
⇒ 16x − 64 − x² − 4x = 48
⇒ − x² − 20x − 64 − 48 = 0
⇒ − x² − 20x − 112 = 0
⇒ x² + 20x + 112 = 0 
Here, a = 1, b = 20 and c = 112
∴ b² − 4ac = (20)² − 4 (1) (112)
= 400 − 448
= − 48 < 0
Since b² − 4ac is less than 0.
∴ The quadratic equation (1) has no real roots.
Thus, the given equation is not possible.

Question. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Answer : Let the breadth of the rectangle be x metres.
Since, the parimetre of the rectangle = 80 m.
∴ 2 [Length + Breadth] = 80
2 [Length + x] = 80
⇒ Length + x = 80/2 = 40
⇒ Length = (40 − x) metres
∴ Area of the rectangle = Length × breadth
= (40 − x) × x sq. m
= 40x − x2
Now, according to the given condition,
Area of the rectangle = 400 m²
∴ 40x − x² = 400
⇒ − x² + 40x − 400 = 0
⇒ x² − 40x + 400 = 0 
Comparing (1) with ax² + bx + c = 0, we get
a = 1
b = − 40
c = 400
∴ b² − 4ac = (− 40)² − 4 (1) (400)
= 1600 − 1600 = 0
Thus, the equation (1) has two equal and real roots.
∴ x = − b/2a and x = − b/2a
∴ breadth = − (−40)/2(1) = 40/2 = 20 
∴ Breadth, x = 20 m
∴ Length = (40 − x) = (40 − 20) m = 20 m.

Please click on below link to download CBSE Class 10 Mathematics Quadratic Equations Worksheet Set B