# Important Questions Class 10 Maths Chapter 4 Quadratic Equations

In this article explore, Important Questions Class 10 Maths Chapter 4 Quadratic Equations. Quadratic equations have a very long history. Often ancient Indian mathematicians like Shri Brahmagupta(C.E.598â€“665) and Shri Sridharacharya (C.E. 1025) are given credit for developing it. It is the method of completing the square for solving a polynomial equation. These equations are useful in solving our day-to-day problems as well as complex activities such as launching a satellite. A basic quadratic equation can be written as Ax2 By C = 0 here â€˜Aâ€™, â€˜Bâ€™, and â€˜Câ€™ is the constant term, â€˜Ax2â€² is the quadratic term, â€˜Byâ€™ is the linear term and â€˜Aâ€™ should not be zero.

Students preparing for important questions class 10 maths Chapter 4 Quadratic Equations should pay attention to the following topics:

- Finding roots of quadratic equations with the factorisation method
- Finding roots of quadratic equations by completing the square method
- Quadratic formula
- Distinct real roots in quadratic equations

## Important Questions Chapter 4 – Quadratic Equations

**Q1. Find the value of p so that the quadratic equation px(x â€“ 3) + 9 = 0 has two equal roots.Â **

**Solution:**

We have, px (x â€“ 3) + 9 = 0

px2 â€“ 3px + 9 = 0 Here a = p, b = -3p,

D = 0

b2 â€“ 4ac = 0 â‡’ (-3p)2 â€“ 4(p)(9) = 0

â‡’ 9p2 â€“ 36p = 0

â‡’ 9p (p â€“ 4) = 0

â‡’ 9p = 0 or p â€“ 4= 0

p = 0 (rejected) or p = 4

âˆ´ p = 4 â€¦â€¦..(âˆµ Coeff. of x2 cannot be zero)

**Q2. Find the value of m so that the quadratic equation mx (x â€“ 7) + 49 = 0 has two equal roots.Â **

**Solution:**

We have, mx (x â€“ 7) + 49 = 0

mx^2 â€“ 7mx + 49 = 0

Here, a = m, b = â€“ 7m, c = 49

D = b^2 â€“ 4ac = 0 â€¦[For equal roots

â‡’ (-7m)^2 â€“ 4(m) (49) = 0

â‡’ 49m^2 â€“ 4m (49) = 0

â‡’ 49m (m â€“ 4) = 0

â‡’ 49m = 0 or m â€“ 4 = 0

m = 0 (rejected) or m = 4

âˆ´ m = 4

**Q3. Solve the following quadratic equation for x: x^2 â€“ 2ax â€“ (4b^2 â€“ a^2) = 0**

**Solution:**

Given quadratic equation can be written as

x^2 â€“ 2ax â€“ 4b^2 + a^2 = 0.

(x^2 â€“ 2ax + a^2) â€“ 4b^2 = 0 or (x â€“ a)^2 â€“ (2b)^2 = 0

As we know,

[a^2 â€“ b^2 = (a + b)(a â€“ b)]

âˆ´ (x â€“ a + 2b) (x â€“ a â€“ 2b) = 0

â‡’ x â€“ a + 2b = 0 or x â€“ a â€“ 2b = 0

â‡’ x = a â€“ 2b or x = a + 2b

â‡’ x = a â€“ 2b and x = a + 2b

**Q4. Find the value(s) of k so that the quadratic equation 3x^2 â€“ 2kx + 12 = 0 has equal roots.Â **

**Solution:**

Given: 3x^2 â€“ 2kx + 12 = 0

Here a = 3, b = -2k, c = 12

D = 0 â€¦ [Since roots are equal As

b2 â€“ 4ac = 0

âˆ´ (-2k)^2 â€“ 4(3) (12) = 0

â‡’ 4k^2 â€“ 144 = 0 â‡’ k^2 = 144/4 = 36

âˆ´ k = Â±âˆš36=Â±6

**Q5. For what value of k, are the roots of the quadratic equation y^2 + k^2 = 2 (k + 1)y equal?Â **

**Solution:**

y^2 + k^2 = 2(k + 1)y

y^2 â€“ 2(k + 1)y + k^2 = 0

Here a = 1, b = -2(k + 1), c = k^2

D = 0 â€¦ [Roots are equal

b^2 â€“ 4ac = 0

âˆ´ [-2(k + 1)]2 â€“ 4 Ã— (1) Ã— (k2) = 0

â‡’ 4(k2 + 2k + 1) â€“ 4k2 = 0

â‡’ 4k2 + 8k + 4 â€“ 4k2 = 0

â‡’ 8k + 4 = 0

â‡’ 8k = -4 âˆ´ k = âˆ’48=âˆ’12

**Q6. ****Find that value of p for which the quadratic equation (p + 1)x^2 â€“ 6(p + 1)x + 3 (p + 9) = 0, p â‰ -1 had equal roots.Â **

**Solution:**

For the given quadratic equation to have equal roots, D = 0

Here a = (p + 1), b = -6(p + 1), c = 3(p + 9)

D = b^2 â€“ 4ac

â‡’ [-6(p + 1)]^2 â€“ 4(p + 1).3 (p + 9) = 0

â‡’ 36(p + 1)^2 â€“ 12(p + 1) (p + 9) = 0

â‡’ 12(p + 1) (3p + 3 â€“ p â€“ 9) = 0

â‡’ 12(p + 1)(2p â€“ 6) = 0

â‡’ 24(p + 1)(p â€“ 3) = 0

â‡’ p + 1 = 0 or p â€“ 3 = 0

â‡’ p = -1 (rejected) or p = 3

âˆ´ p = 3

**Q7. If the roots of the quadratic equation (a â€“ b)x^2 + (b â€“ c)x + (c â€“ a) = 0 are equal, prove that 2a = b + c**

**Solution:**

Hereâ€™aâ€™ = a â€“ b, â€˜bâ€™ = b â€“ c, â€˜câ€™ = c â€“ a

D = 0 â€¦.[Roots are equal

b^2 â€“ 4ac = 0

â‡’ (b â€“ c)^2 â€“ 4(a â€“ b)(c â€“ a) = 0

â‡’ b^2 + c^2 â€“ 2bc â€“ 4(ac â€“ a^2 â€“ bc + ab) = 0

â‡’ b^2 + c^2 â€“ 2bc â€“ 4ac + 4a^2 + 4bc â€“ 4ab = 0

â‡’ 4a^2 + b2^2 + c^2 â€“ 4ab + 2bc â€“ 4ac = 0

â‡’ (-2a)^2 + (b)^2 + (c)2^2 + 2(-2a)(b) + 2(b)(c) + 2(c)(-2a) = 0

â‡’ [(-2a) + (b) + (c)]^2 = 0

â€¦.[âˆµ x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2

Taking square-root on both sides

-2a + b + c = 0

â‡’ b + c = 2a âˆ´ 2a = b + c

**Q8. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.**

**Solution:**

Let three consecutive natural numbers are x, x + 1, x + 2.

According to the question,

(x + 1)^2 â€“ [(x + 2)^2 â€“ x^2] = 60

â‡’ x^2 + 2x + 1 â€“ (x^2 + 4x + 4 â€“ x^2) = 60

â‡’ x^2 + 2x + 1 â€“ 4x â€“ 4 â€“ 60 = 0

â‡’ x^2 â€“ 2x â€“ 63 = 0

â‡’ x^2 â€“ 9x + 7x â€“ 63 = 0

â‡’ x(x â€“ 9) + 7(x â€“ 9) = 0

â‡’ (x â€“ 9) (x + 7) = 0

â‡’ x â€“ 9 = 0 or x + 7 = 0

â‡’ x = 9 or x = -7

Natural nos. can not be -ve, âˆ´ x = 9

âˆ´ Numbers are 9, 10, 11.

**Q9. If the sum of two natural numbers is 8 and their product is 15, find the numbers.**

**Solution:**

Let the numbers be x and (8 â€“ x).

According to the Question,

x(8 â€“ x) = 15

â‡’ 8x â€“ x^2 = 15

â‡’ 0 = x^2 â€“ 8x + 15

â‡’ x^2 â€“ 5x â€“ 3x + 15 = 0

â‡’ x(x â€“ 5) â€“ 3(x â€“ 5) = 0

â‡’ (x â€“ 3)(x â€“ 5) = 0

x â€“ 3 = 0 or x â€“ 5 = 0

x = 3 or x = 5

When x = 3, numbers are 3 and 5.

When x = 5, numbers are 5 and 3.

**Q10. Sum of the areas of two squares is 400 cm^2. If the difference of their perimeters is 16 cm, find the sides of the two squares.**

**Solution:**

Let the side of Large square = x cm

Let the side of small square = y cm

According to the Question,

x^2 + y^2 = 400â€¦ (i) â€¦[âˆµ area of square = (side)^2

4x â€“ 4y = 16 â€¦[âˆµ Perimeter of square = 4 sides

â‡’ x â€“ y = 4 â€¦ [Dividing both sides by 4

â‡’ x = 4 + y â€¦(ii)

Putting the value of x in equation (i),

(4 + y)^2 + y2^2 = 400

â‡’ y2 + 8y + 16 + y^2 â€“ 400 = 0

â‡’ 2y^2 + 8y â€“ 384 = 0

â‡’ y^2 + 4y â€“ 192 = 0 â€¦ [Dividing both sides by 2

â‡’ y^2 + 16y â€“ 12y â€“ 192 = 0

â‡’ y(y + 16) â€“ 12(y + 16) = 0

â‡’ (y â€“ 12)(y + 16) = 0

â‡’ y â€“ 12 = 0 or y + 16 = 0

â‡’ y = 12 or y = -16 â€¦ [Neglecting negative value

âˆ´ Side of small square = y = 12 cm

and Side of large square = x = 4 + 12 = 16 cm

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