## Solutions of adfected Quadratic equations – Quadratic Equations:

We know the general form of an adfected quadratic equation is ax² + bx + c, a≠0. this equation can also occur in different forms such as ax² + bx = 0, ax² + c = 0, and ax² = 0.

### Solution of a quadratic equation by factorisation method – Quadratic Equations:

Factorisation method is used when the quadratic equation can be factorised can be factorised into two linear factors. After factorisation, the quadratic equation is expressed as the product of its two linear factors and this is equated to zero. That is ax² + bx + c = 0 and (x ± m)(x ± n) = 0. Then we apply zero product rule and equate each factor to zero and solve for the unknown.

i.e.,

So, ±m and ±n are the roots of the quadratic equation ax² + bx + c = 0.

## Quadratic Equations – Exercise 9.3

#### Solve the quadratic equations by factorization method:

(i) x^{2} + 15x + 50 = 0

(ii) 6 – p^{2} = p

(iii) 100x^{2} – 20x + 1 = 0

(iv) √2x^{2} + 7x +5√2 = 0

(v)x^{2} + 4kx + 4k^{2} = 0

(vi) m – ^{7}/_{m} = 6

(vii) 0.2t^{2} – 0.04t = 0.03

(viii) √5x^{2} + 2x= 3√5

(ix) ^{x}/_{x+1 }+ ^{x+1}/_{x} = ^{34}/_{15}

(x) ^{x-1}/_{x-2} + ^{x-3}/_{x-4} = 3^{1}/_{3}

(xi) a^{2}b^{2}x^{2} –(a^{2} + b^{2})x + 1 = 0

(xii) (2x – 3) = √(2x^{2} – 2x + 2)

## Quadratic Equations – Exercise 9.3

**Solve the quadratic equations by factorisation method:**

**(i) x ^{2} + 15x + 50 = 0**

Solution:

x^{2} + 15x + 50 = 0

x^{2} + 10x + 5x + 50 = 0

x(x + 10) + 5(x + 10) = 0

(x + 10)(x + 5) = 0

x = -10 or x = -5

** (ii) 6 – p ^{2} = p**

Solution:

6 – p^{2} – p = 0

p^{2} + p – 6 = 0

p^{2} + 3p – 2p – 6 = 0

p(p + 3)-2(p + 3) = 0

(p – 2)(p + 3) = 0

p = 2 or p = -3

** (iii) 100x ^{2} – 20x + 1 = 0**

Solution:

(10x)^{2} – 20x + 1 = 0

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1)-1(10x – 1) = 0

(10x – 1)(10x – 1) = 0

10x = 1 or 10x = 1

x = ^{1}/_{10} or x = ^{1}/_{10}

** (iv) √2x ^{2} + 7x +5√2 = 0**

Solution:

√2x^{2} + 7x +5√2 = 0

√2x^{2} + 2x + 5x +5√2 = 0

√2x(x + √2) + 5(x + √2) = 0

(√2x + 5)(x + √2) = 0

√2x = -5 or x = -√2

x = ^{-5}/_{√2} or x = -√2

** (v)x ^{2} + 4kx + 4k^{2} = 0**

Solution:

x^{2} + 4kx + 4k^{2} = 0

x^{2} + 2kx + 2kx + 4k^{2} = 0

x(x + 2k)+2k(x + 2k) = 0

(x + 2k)(2 + 2k) = 0

x = -2k or x = -2k

** (vi) m – ^{7}/_{m} = 6**

Solution:

m – ^{7}/_{m} = 6

m^{2} – 7 = 6m

m^{2} – 6m – 7 = 0

m^{2} – 7m + m – 7 = 0

m(m – 7) +1(m – 7) = 0

(m + 1)(m – 7) = 0

m = -1 or m = 7

** (vii) 0.2t ^{2} – 0.04t = 0.03**

Solution:

0.2t^{2} – 0.04t = 0.03

0.2t^{2} – 0.04t – 0.03 = 0

20t^{2} – 4t – 3 = 0

20t^{2} – 10t + 6t – 3 = 0

10t(2t – 1)+3(2t – 1) = 0

(10t + 3)(2t – 1) = 0

10t + 3 = 0 or 2t – 1 = 0

10t = -3 or 2t = 1

t = ^{-3}/_{10} or t = ^{1}/_{2}

** (viii) √5x ^{2} + 2x= 3√5**

Solution:

√5x^{2} + 2x= 3√5

√5x^{2} + 2x – 3√5 = 0

√5x^{2} + 3x – 5x – 3√5 = 0

x(√5x + 3) -√5(√5x + 3) = 0

(x – √5)( √5x + 3) = 0

x – √5 = 0 or √5x + 3 = 0

x = √5 or √5x = – 3

x = √5 or x = ^{-3}/_{√5}

** (ix) ^{x}/_{x+1 }+ ^{x+1}/_{x} = ^{34}/_{15}**

Solution:

^{x}/_{x+1 }+ ^{x+1}/_{x} = ^{34}/_{15}

^{x}/_{x+1 }+ ^{x+1}/_{x} – ^{34}/_{15} = 0

^{x[15x(x+1)]}/_{x+1 }+ ^{(x+1)[15x(x+1)]}/_{x} – ^{34[15x(x+1)]}/_{15} = 0

15x^{2} + 15(x + 1)^{2} – 34x(x+1) = 0

15x^{2} + 15(x^{2} + 2x + 1) – 34x^{2} – 34x = 0

15x^{2} + 15x^{2} +30x + 15 – 34x^{2} – 34x = 0

-4x^{2} – 4x +15 = 0

4x^{2} + 4x – 15 = 0

4x^{2} – 6x + 10x – 15 = 0

2x(2x – 3) +5(2x – 3) = 0

(2x + 5)(2x – 3) = 0

2x + 5 = 0 or 2x – 3 = 0

2x = -5 or 2x = 3

x = –^{5}/_{2} or x = ^{3}/_{2}

** (x) ^{x-1}/_{x-2} + ^{x-3}/_{x-4} = 3^{1}/_{3}**

Solution:

^{x-1}/_{x-2} + ^{x-3}/_{x-4} = 3^{1}/_{3}

^{x-1}/_{x-2} + ^{x-3}/_{x-4} – ^{10}/_{3} = 0

^{(x-1)(x – 2)(x – 4)3}/_{(x-2)} + ^{(x – 3)(x – 2)(x – 4)3}/_{(x – 4)} – ^{10(x – 2)(x – 4)3}/_{3} = 0

3(x – 1)(x – 4) + 3(x – 3)(x – 2) – 10(x – 2)(x – 4) = 0

3(x^{2} – 4x – x + 4) + 3(x^{2} – 2x – 3x + 6) – 10(x^{2} – 4x – 2x + 8) = 0

3x^{2} – 12x – 3x + 12 + 3x^{2} – 6x – 9x + 18 – 10x^{2} + 40x + 20x – 80 = 0

-4x^{2} + 30x – 50 = 0

4x^{2} – 30x + 50 = 0

2x^{2} – 15x + 25 = 0

2x^{2} – 10x – 5x + 25 = 0

2x(x – 5)-5(x – 5) = 0

(2x – 5)(x – 5) = 0

2x – 5 = 0 or x – 5 = 0

x = ^{5}/_{2} or x = 5

** (xi) a ^{2}b^{2}x^{2} –(a^{2} + b^{2})x + 1 = 0**

Solution:

a^{2}b^{2}x^{2} –(a^{2} + b^{2})x + 1 = 0

a^{2}b^{2}x^{2} –a^{2}x – b^{2}x + 1 = 0

a^{2}x(b^{2}x – 1)-1(b^{2} x – 1) = 0

(a^{2}x – 1)(b^{2}x – 1) = 0

a^{2}x – 1 = 0 or b^{2}x – 1 = 0

a^{2}x = 1 or b^{2}x = 1

x = ^{1}/_{a^2} or x = ^{1}/_{b^2}

** (xii) (2x – 3) = √(2x ^{2} – 2x + 21)**

Solution:

(2x – 3) = √(2x^{2} – 2x + 21)

(2x – 3)^{2} = (2x^{2} – 2x + 21)

4x^{2} – 12x + 9 = 2x^{2} – 2x + 21

4x^{2} – 12x + 9 – (2x^{2} – 2x + 21) = 0

4x^{2} – 12x + 9 – 2x^{2} + 2x – 21 = 0

2x^{2} – 10x – 12 = 0

x^{2} – 5x – 6 = 0

x^{2} – 6x + x – 6 = 0

x(x – 6) +1(x – 6) = 0

(x – 6)(x + 1) = 0

x – 6 = 0 or x + 1 = 0

x = 6 or x = -1