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Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

SHORT ANSWER TYPE QUESTIONS

Q1. Solve 2x2 − 5x + 3 = 0.

Sol. We have:

2x2 − 5x + 3 = 0     ...(1)

Comparing (1) with ax2 + bx + c = 0,

∴ a = 2
b = − 5
c = 3

∴ b2 − 4ac =(− 5)2 − 4 (2) (3)
25 − 24 = 1
Since, Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking, + ve sign,
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Taking, −ve sign,

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Thus, the required roots are
x = 3/2 and x = 1

Q2. Solve the following quadratic equation: 

x2 + 4x − 8 = 0
Sol. We have:

2x+ 4x − 8= 0

Dividing by 2, we get

x2 + 2x − 4 = 0 ...(1)

Comparing (1) with ax2 + bx + c = 0,

a = 1
b = 2
c = − 4

∴ b− 4ac = (2)2 − 4 (1) (− 4)
= 4 + 16 = 20
Since, Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Taking +ve sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking −ve sign we get,

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Thus, the required roots are x =  Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol. We have,
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ x2 + 3x + 2 + x2 − 3x + 2 =3 (x2 + x − 2)
⇒ 2x2 + 4 = 3x2 + 3x − 6
⇒ 3x2 + 3x − 6 − 2x2 − 4=0
⇒ 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

Q4. Solve (using quadratic formula): 

x2 + 5x + 5 = 0

Sol. 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, Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking +ve sign, we have:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking −ve sign, we have:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Thus, the required roots are:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q5. Solve for x:  36x2 − 12ax + (a2 − b2) = 0.

Sol. We have:

36x2 − 12ax + (a2 − b2) = 0 ...(1)

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

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Taking +ve sign, we have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking −ve sign, we get
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Thus, the required roots are:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q6. Find the roots of the quadratic equation Class 10 Maths Chapter 4 Question Answers - Quadratic Equations using the quadratic formula.

Sol. Comparing the given equation with the general equation ax2 + bx + c = 0, we have

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

= 5 + 16
= 21

Now, using the quadratic formula, we have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the, positive sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the negative sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q7. Solve: 16x2 − 8a2 x + (a4 − b4) = 0 for x.

Sol. We have:

16x2 − 8a2 x + a4 − b4 = 0 ...(1)
Comparing (1) with ax+ 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

Since, Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Now, taking +ve sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking −ve sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Thus, the required roots are:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q8. Solve for x:  9x2 − 6ax + a2 − b2 = 0.

Sol. We have:

9x2 − 6ax + (a− b2) = 0     ...(1)
Comparing (1) with ax2 + bx + c = 0, we get

a = 9, b = − 6a and c = (a2 − b2)
∴ b2 − 4ac = (− 6a)− 4 (9) (a2 − b2)
= 36a2 − 36 (a2 − b2)
= 36a2 − 36a2 + 36b2 
= 36b2 = (6b)2

Since, Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the +ve sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the −ve sign, we get

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

∴ The required roots are:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q9. Evaluate Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol. Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

The given expression can be written as
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
or Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

squaring both side, we have
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
⇒ x2 = 20 + x
⇒ x2 – x – 20 = 0, where
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Here : a =1,  b = –1  and  c = –20

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Since the given expression is positive,
∴ Rejecting the negative sign, we have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q10. Using quadratic formula, solve the following quadratic equation for x:

x2 − 4ax + 4a2 − b2 = 0

Sol. 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, Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

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

Q11. Using quadratic formula, solve the following quadratic equation for x: 

x2 − 2ax + (a2 − b2) = 0

Sol. Comparing the given equation with ax2 + bx + c = 0, we have:

a = 1, b = − 2a and c = (a2 − b2)
∴ b2 − 4ac =(− 2a)2 − 4 (1) (a2 − b2)
= 4a2 − 4 (a2 − b2)
= 4a2 − 4a2 + 4b2 = 4b2

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the +ve sign, we get

x = a + b

Taking the −ve sign, we get

x = a − b

Thus, the required roots are:

x = a + b and x = a − b

Q12. Solve for x : 

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol. We have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
⇒ 2 × 3  = 2 (x − 1) (x − 3)
⇒ 3 = (x − 1) (x − 3)
⇒ x2 − 4x + 3 − 3 = 0
⇒ x2 − 4x = 0
⇒ x(x − 4) = 0
Either x = 0 or  x = 4
Thus,   x = 0 ; 4

Q13. Find the roots of the equation:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Sol. We have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ (3x + 2) (7x + 9) = 11 (2x2 + 5x − 3)
⇒ 21x2 + 27x + 14x + 18 = 22x2 + 55x − 33
⇒ 21x2 + 41x + 18 = 22x2 + 55x − 33
⇒ (21 − 22) x2 + (41 − 55)x + (18 + 33) = 0
⇒ − x2 + (− 14x) + (51) = 0
⇒ x2 + 14x − 51 = 0
⇒ x2 + 17x − 3x − 51 = 0
⇒ x (x + 17) − 3 (x + 17) = 0
⇒ (x + 17) (x − 3) = 0    

Either x + 17 =0 ⇒ x = − 17
or x − 3 = 0 ⇒ x = 3

Thus, the required roots of the given equation are:

3 and −17

Q14. Find the roots of the equation:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Sol. We have:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
⇒ Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ − 3 × 3 = 4 × (x2 − 3x)
⇒ − 9 = 4x2 − 12x
⇒ 4x2 − 12x + 9 = 0 ...(1)

Comparing (1), with ax2 + bx + c = 0, we get

a = 4
b = − 12
c = 9

∴ b2 − 4ac =(− 12)2 − 4 (4) (9)
= 144 − 144 = 0

Now, the roots are:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
⇒ x = 12/8 = 3/2

Thus, the roots are: 3/2 and 3/2

Q15. Solve for x :

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol.
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ 3 (x + 2) =(x − 2) (x + 1) or  3x + 6    
= x2 − x − 2

⇒ x2 − 4x − 8 = 0
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Q16. Find the roots of the equation:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol. We have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
⇒ Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ (2x − 2) (2x + 5) = 8 (x2 − 2x)
⇒ 4x2 + 10x − 4x − 10 = 8x2 − 16x
⇒− 4x2 + 22x − 10 = 0
⇒ 2x2 − 11x + 5 = 0   ...(1)

Comparing (1) with ax2 + bx + c = 0, we have:

a = 2
b = − 11
c = 5
∴ b2 − 4ac = (− 11)2 − 4 (2) (5)
= 121 − 40 = 81

Now, the roots are given by
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the +ve sign,

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Taking the −ve sign,

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Thus, the required roots are: 5 and 1/2.

Q17. Solve : 

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol. We have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ x (a + b + x) = −ab
⇒ x+ (a + b)x + ab = 0
⇒ x + ax + bx + ab = 0
⇒ x (x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
⇒ x + a = 0   or  x + b = 0
∴ x = −a  or   x = −b

Q18. Find the roots of the following equation:

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Sol. We have:
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

⇒ − 11 × 30 = 11 × (x2 − 3x − 28)
⇒ − 30 = x2 − 3x − 28
⇒ x2 − 3x − 28 + 30 = 0
⇒ x2 − 3x + 2 = 0
⇒ x2 − 2x − x + 2 = 0
⇒ x (x − 2) − 1 (x − 2)= 0
⇒ (x − 1) (x − 2) = 0

Either x −1= 0 ⇒ x = 1

or x −2 = 0 ⇒ x = 2

Thus, the required roots are: 1 and 2.

Q19. If α and β are roots of the equation x2 – 1 = 0, form an equation whose roots are  Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Sol. ∵ α and β are roots of x– 1 = 0 and x2 – 1 = 0 can be written as x2 + 0x – 1= 0 where a = 1, b = 0 and c = –1.

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

Now, the roots of the new equation areClass 10 Maths Chapter 4 Question Answers - Quadratic Equations

∴ Sum of the roots of the new equation
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Class 10 Maths Chapter 4 Question Answers - Quadratic Equations 
Product of the roots of the new equation

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations
Since, a quadratic equation is given by

Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

∴ The required quadratic equation is

x2 – (–4) x + 4 = 0
or x2 + 4x + 4 = 0

Remember α2 + β2 + 2αβ =(α + β)2
⇒ α2 + β2 =(α + β)2 – 2αβ

The document Class 10 Maths Chapter 4 Question Answers - Quadratic Equations is a part of the SSS 1 Course Mathematics for SSS 1.
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FAQs on Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of the second degree, which means it has a variable raised to the power of 2. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants.
2. How do you solve a quadratic equation?
Ans. Quadratic equations can be solved using various methods such as factoring, completing the square, or using the quadratic formula. The most commonly used method is the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac)) / (2a).
3. Can a quadratic equation have more than two solutions?
Ans. No, a quadratic equation can have at most two solutions. This is because a quadratic equation represents a parabola, which can intersect the x-axis at most two times. However, it is also possible for a quadratic equation to have only one solution or no real solutions, depending on the discriminant (b^2 - 4ac).
4. What is the discriminant of a quadratic equation?
Ans. The discriminant of a quadratic equation is the part of the quadratic formula that appears under the square root sign, i.e., b^2 - 4ac. It helps determine the nature of the solutions. If the discriminant is positive, the equation has two distinct real solutions. If the discriminant is zero, the equation has one real solution (also known as a repeated root). And if the discriminant is negative, the equation has no real solutions, only complex solutions.
5. Can a quadratic equation have no solutions?
Ans. Yes, a quadratic equation can have no real solutions if the discriminant (b^2 - 4ac) is negative. In this case, the solutions will be complex numbers. For example, the equation x^2 + 1 = 0 has no real solutions but has two complex solutions: x = ±i, where i is the imaginary unit.
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