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Topic– 2: Solution of a Quadratic Equations by Factorisation

Very Short Answer Type Questions: [1 Mark]

1. Find the roots of the quadratic equation x2 + 7x + 12 = 0.

Sol. x2 + 7x + 12 = 0

⇒ x2 + 3x + 4x + 12 = 0

⇒ x(x + 3) + 4(x + 3) = 0

⇒ (x + 3)(x + 4) = 0

⇒ x = –3, – 4

2. Find the roots of the equation x2 – √3 x – x + √3 = 0.

Sol. x2 – √3 x – x + √3 = 0

⇒ x(x – √3 ) – 1(x – √3 ) = 0

⇒ (x – √3 )(x – 1) = 0 ⇒ x = √3 , 1

3. Find the roots of the quadratic equation √3 x2 – 2x – √3 = 0.

Sol. √3 x2 – 2x – 3 = 0

⇒ √3 x2 – √3x + x – √3 = 0

⇒ √3 x(x – √3 ) + 1(x – √3 ) = 0

⇒ (x – √3 )( √3 x + 1) = 0 ⇒ x = √3 , 1/√3

5. Find the roots of x2x – a(a + 1) = 0.

Sol. We have, x2 – x – a(a + 1) = 0

⇒ x2 – (a + 1) x + ax – a (a + 1) = 0

⇒ x{x – (a + 1)} + a{x –(a + 1)} = 0

⇒ [x – (a + 1)] (x + a) = 0

⇒ x = a + 1 or x = –a

6. If is a root of the equation x2 + kx – 5/4 = 0. Then find the value of k.

Sol. What we have is a roof of equations. Then,

= 2k – 4 = 0 ⇒ 2k = 4 ⇒ k = 2.

7. Find the value of m for which the given equations 12x2 + 4mx + 3 = 0 has equal roots.

Sol. We have,

a = 12, b = 4m and c = 3

For equal roots, D = 0.

⇒ b2 – 4 ac = 0

⇒ (4m)2 – 4× 12× 3 = 0

⇒ 16m2 – 144 = 0

⇒ 16m2 = 144

⇒ m2 = 144/19 = 9

⇒ m = ± 3 .

8. Find the roots of the equation x2 + x – p (p + 1) = 0, where p is a constant.

Sol. x2 + x – p (p + 1) = 0

⇒ x2 + x – p2 – p = 0

⇒ x2 – p2 + x – p = 0

⇒ (x + p) (x – p) + (x – p) = 0

⇒ (x – p) (x + p + 1) = 0

⇒ x – p = 0

⇒ x =p

or x + p + 1 = 0

or x = – (p + 1).

9. Find the roots of the equation x2 – 3x – m(m + 3) = 0, where m is a constant.

Sol. We have,

x2 – 3 x – m(m + 3) = 0

⇒ x2 – 3 x – m2 – 3m = 0

⇒ x2 – m2 – 3 x – 3m = 0

⇒ (x + m) (x – m) – 3 (x + m) = 0

⇒ (x + m) [x – m – 3 ] = 0

⇒ x + m = 0

⇒ x = –m

or x – m – 3 = 0

or x = m + 3

10. If 1 is a root of the equation ay2 + ay + 3 and y2 + y + b = 0, then find the value of ab.

Sol. ay2 + ay + 3 = 0

⇒ a(1)2 + a(1) + 3 = 0

⇒ a + a + 3 = 0

⇒ 2a = –3

⇒ 2a = –3

⇒ a = –2/3

Also, y2 + y + b = 0

⇒ (1)2 + (1) + b = 0

⇒ 1 + 1 + b = 0

⇒ b = – 2

Now,

11. If the quadratic equation px2 –2√5px + 15 = 0, has two equal roots then find the value of p.

Sol. If the given equation has two equal roots, then D = 0.

⇒ b2 – 4 ac = 0

⇒ (–2√5px) – 4(p) + (15) = 0

⇒ 20p2 – 60p = 0

⇒ 20p (p – 3) = 0

⇒ 20p = 0 or p – 3 = 0

⇒ p = 0

or p = 3

But p = 0 is not possible.

∴ p = 3.

12. If x = –1/2 , is a solution of the quadratic equation 3x2 + 2kx – 3 = 0, find the vlaue of k.

Sol. We have, 3x2 + 2kx – 3 = 0

Short Answer Type Questions: [2 Marks]

1. Find the quadratic equation whose roots are (2 +2√3) and (2 –2 √3).

Sol. x2 – 4x – 8 = 0

So, they are roots of x2 – 4x – 8 = 0

2. Find the roots of a2x2 + (a2 – b2)x – b2 = 0.

Sol. We have, a2x2 + (a2 – b2)x – b2 = 0

⇒ a2x (x + 1) – b2(x + 1) = 0

⇒ (a2x – b2) (x + 1) = 0

∴ x = –1 and x = b2a

3. When k = 3, find the roots of (4 – k) x2 + (2k + 4) x + (8k + 1) = 0

Sol. When k = 3,

(4 – k)x2 + (2k + 4)x + (8k + 1) = 0

⇒ x2 + 10x + 25 = 0

⇒ (x + 5) (x + 5) = 0 ⇒ x = –5, –5

4. Find the roots of the equation

3a2x2 + 8abx + 4b2 = 0.

Sol. We have, 3a2x2 + 8abx + 4b2 = 0

⇒ 3a2x2 + 6abx + 2abx + 4b2 = 0

⇒ 3ax(ax + 2b) + 2b(ax + 2b) = 0

⇒ (3ax + 2b) (ax + 2b) = 0

∴ x = −2b/ a or x = −2 b/ 3a

5. Find the quadratic equation where roots are x = a +b/ 2 and x = a − b / 2 .

Sol. 4x2 – 4ax + (a2 – b2) = 0

⇒ 4x2 – 2(a + b)x – 2(a – b)x + (a2 – b2) = 0

⇒ 2x{2x – (a + b)} – (a – b)

{2x – (a + b)}{2x – (a – b)}

6. Solve for x : abx2 + (b2 – ac)x – bc = 0

Sol. We have, abx2 + (b2 – ac)x – bc = 0

⇒ abx2 + b2x – acx – bc = 0

⇒ bx(ax + b) – c(ax + b) = 0

⇒ (ax + b) (bx – c) = 0

⇒ ax + b = 0 or bx – c = 0

⇒ ax = – b or bx = c

x = − b/a or x = c/b

7. Solve the following quadratic equation for x :

p2x2 + (p2 – q2)x – q2 = 0

Sol. We have, p2x2 + (p2 – q2)x – q2 = 0

⇒ p2x2 + p2x – q2x – q2 = 0

⇒ (p2x2 + p2x) – (q2x + q2) = 0

⇒ p2x(x + 1) – q2(x + 1) = 0

⇒ (x + 1)(p2x – q2) = 0

⇒ x + 1 = 0 or, p2x – q2 = 0

⇒ x = –1 or, x = q2/p2

1. Find the roots of the following equation:

Sol. We have,

⇒   x2 – 3x – 28 = –30

⇒ x2 – 3x + 2 = 0

⇒ x2 – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2) (x – 1) = 0

⇒ x = 2 or x = 1

2. Solve the following quadratic equations by factorisation method:

⇒   34x2 + 34x = 15x2 + 15x2 + 30x + 15

⇒ 4x2 + 4x – 15 = 0

⇒ 4x2 + 10x – 6x – 15 = 0

⇒ 2x(2x + 5) – 3(2x + 5) = 0

⇒ (2x + 5) (2x – 3) = 0

⇒ 2x + 5 = 0 or 2x – 3 = 0

⇒ x = – 5/2 or x = 3/2

⇒   4x2 – 4x + 2 = 5x2 – 5x

⇒ x2 – x – 2 = 0

⇒ x2 – 2x + x – 2 = 0

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x – 2) (x + 1) = 0

⇒ x = 2 or x = –1

⇒   8x2 + 8 = 17x2 – 34x

⇒ 9x2 – 34x – 8 = 0

⇒ 9x2 – 36x + 2x – 8 = 0

⇒ 9x(x – 4) + 2(x – 4) = 0

⇒ (x – 4) (9x + 2) = 0

⇒ x – 4 = 0 or 9x + 2 = 0

⇒ x = 4 or x = − 2/9 .

3. Solve the following quadratic equations by factorisation method :

⇒  (4 – 3x) (2x + 3) = 5x

⇒ 12 – x – 6x2 = 5x

⇒ 6x2 + 6x – 12 = 0

⇒ x2 + x – 2 = 0

⇒ x2 + 2x – x – 2 = 0

⇒ x(x + 2) – 1(x + 2) = 0

⇒ (x + 2) (x – 1) = 0

⇒ x + 2 = 0 or x – 1 = 0

x = –2 or x = 1.

(ii) Clearly, the given equation is valid if x – 3 ≠ 0 and 2x + 3 ≠ 0, i.e.,

when x ≠ – 3/2 , 3 Now,

⇒   2x(2x + 3) + (x – 3) + 3x + 9 = 0 [Multiplying throughout by (x – 3)(2x + 3)]

⇒ 4x2 + 6x + x – 3 + 3x + 9 = 0

⇒ 4x2 + 10x + 6 = 0

⇒ 2x2 + 5x + 3 = 0

⇒ 2x2 + 2x + 3x + 3 = 0

⇒ 2x(x + 1) + 3(x + 1) = 0

⇒ (x + 1) (2x + 3) = 0

⇒ x + 1 = 0 [∵   2x + 3 ≠ 0]

⇒ x = –1

Hence, x = –1 is the only solution of the given equation.

⇒   (x – 1)(x – 4) = 18 ⇒ x2 – 5x + 4 = 18

⇒ x2 – 5x – 14 = 0

⇒ x2 – 7x + 2x – 14 = 0

⇒ x(x – 7) + 2(x – 7) = 0

⇒ (x – 7)(x + 2) = 0

⇒ x – 7 = 0 or x + 2 = 0

⇒ x = 7 or x = –2.

⇒   x(a + b + x) = – ab

⇒ x2 + 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

Hence, –a and –b are the roots of the given equation.

5. Solve for x :

12abx2 – (9a2 – 8b2)x – 6ab = 0

Sol. We have, 12abx2 – (9a2 – 8b2)x – 6ab = 0

⇒ 12abx2 – 9a2x + 8b2x – 6ab = 0

⇒ 3ax(4bx – 3a) + 2b(4bx – 3a) = 0

⇒ (4bx – 3a)(3ax + 2b) = 0

⇒ 4bx – 3a = 0 or 3ax + 2b = 0

⇒ 4bx = 3a or 3ax = –2b

⇒ x = 3a/4b or x = −2 b/3a

6. Solve for x : 6x2 – 12ax + (a2 – b2) = 0

Sol. We have, 36x2 – 12ax + (a2 – b2) = 0

⇒ (36x2 – 12ax + a2) – b2 = 0

⇒ (6x – a)2 – b2 = 0

⇒ (6x – a + b)(6x – a – b) = 0

⇒ 6x – a + b = 0 or 6x – a – b = 0

⇒ 6x = (a – b) or 6x = (a + b)

⇒ x = a− b/6 or x = a + b/6

7. Find the roots of the equation

x2 – 2 (a2+b2)x + (a2 – b2)2 = 0

Sol. We have x2 – 2 (a2 + b2)x + (a2 – b2)2 = 0

⇒ x2 – {(a + b)2 + (a – b)2} x + (a2 – b2)2 = 0

⇒ x2 – (a + b)2 x – (a – b)2 x + (a2 – b2)2 = 0

⇒ x {x – (a + b)2} – (a – b)2 {x – (a + b)2} = 0

⇒ {x – (a + b)2} {x – (a – b)2} = 0

⇒ x – (a + b)2 = 0

or x – (a – b)2 = 0

⇒ x = (a + b)2 or x = (a – b)2

Long Answer Type questions: [4 Marks]

⇒   (a + b)x2 – 2abx – c(a + b)x + 2abc

= 2cx2 – 2c(a + b)x + 2abc

⇒ x2(a + b – 2c) – x(2ab + ac + bc – 2ac – 2bc) = 0

⇒ x2(a + b – 2c) – x(2ab – ac – bc) = 0

⇒ x{x(a + b – 2c) – (2ab – ac – bc)} = 0

2. Solve : 4x2 – 4a2 x + (a4 – b4) = 0

Sol. We have, 4x2 – 4a2x + (a4 – b4) = 0

Here, constant term

= a4 – b4 = (a2 – b2) (a2 + b2) and,

coefficient of middle term = – 4a2

= – {2(a2 + b2) + 2(a2 – b2)}

∴ 4x2 – 4a2x + (a4 – b4) = 0

⇒ 4x2 – {2(a2 + b2) + 2(a2 – b2)}x + (a2 – b2)(a2 + b2) = 0

⇒ 4x2 – 2(a2 + b2)x – 2(a2 – b2)x + (a2 – b2)(a2 + b2) = 0

⇒ {4x2 – 2(a2 + b2)x} – {2(a2 – b2)x – (a2 – b2)(a2 + b2)} = 0

⇒ 2x {2x – (a2 + b2)} – (a2 – b2) {2x – (a2 + b2) = 0

⇒  {2x – (a2 + b2)}{2x – (a2 – b2)} = 0

⇒ 2x – (a2 + b2) = 0 or 2x – (a2 – b2) = 0

3. Solve : 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0

Sol. Here, constant term = 2a2 + 5ab + 2b2

= 2a2 + 4ab + ab + 2b2

= 2a(a + 2b) + b(a + 2b)

= (a + 2b)(2a + b) and, coefficient of middle term = –9(a + b)

= –3{(2a + b) + (a + 2b)}

∴ 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0

⇒ 9x2 – 3{(2a + b) + (a + 2b)}x + (2a + b)(a + 2b) = 0

⇒ 9x2 – 3(2a + b)x – 3(a + 2b)x

+ (2a + b)(a + 2b) = 0

⇒ 3x{3x – (2a + b)} – (a + 2b){3x – (2a + b)} = 0

⇒ {3x – (2a + b)} {3x – (a + 2b)} = 0

⇒ {3x – (2a + b)} = 0

or {3x – (a + 2b)} = 0

⇒   4y2 + 4 = 10y

⇒ 4y2 – 10y + 4 = 0

⇒ 4y2 – 8y – 2y + 4 = 0

⇒ 4y(y – 2) – 2(y – 2) = 0

⇒ (y – 2) (4y – 2) = 0

⇒ y – 2 = 0 or 4y – 2 = 0

⇒ y = 2 or y = 2/4 = ½

⇒   4x = 2 or 4x = 12 [∵ 4x = y]

⇒ 22x = 21 or 22x = 2–1

⇒ 2x = 1 or 2x = –1

⇒ x = 1/2 or x = – 1/2 .

5. Solve for x :

Sol. We have,

Putting this value in the equation, we get y−10/ y  = 3 ⇒ y 2– 10 = 3y

⇒ y2 – 3y – 10 = 0

⇒ y2 – 5y + 2y – 10 = 0

⇒ y(y – 5) + 2(y – 5) = 0

⇒ (y – 5) (y + 2) = 0

⇒ y = 5 or y = –2

When y = 5, When y = 2,

⇒ 10 x + 5 = 4x – 3

⇒ 6x = –8

⇒ x = − 4/3

⇒ – 4x – 2 = 4x – 3

⇒ – 4x – 4x = –3 + 2

⇒ 8x = 1 ⇒ x = 1/8

⇒  2y2 – 25 = 5y ⇒ 2y2 – 5y – 25 = 0

⇒ 2y2 – 10y + 5y – 25 = 0

⇒ 2y(y – 5) + 5(y – 5) = 0

⇒ (y – 5) (2y + 5) = 0

⇒ y = 5 or y = − 5/2

When y = 5,

⇒ 5x – 15 = 2x + 3

⇒ 5x – 2x = 3 + 15

⇒ 3x = 18 ⇒ x = 6

When y = −5/2

⇒ 4x + 6 = – 5x + 15

⇒ 9x = 9 ⇒ x = 1