Question

Solve the following quadratic equation using factorisation

${4x}^{2} - 4ax + {a}^{2} - {b}^{2} = 0$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: {4x}^{2} - 4ax + {a}^{2} - {b}^{2} = 0$

can be rewritten as

$\rm \: {(2x)}^{2} - 2 \times 2a \times x + {a}^{2} - {b}^{2} = 0$

can be regrouped as

$\rm \: ({(2x)}^{2} - 2 \times 2a \times x + {a}^{2}) - {b}^{2} = 0$

We know,

$\boxed{\sf{  \:\rm \: {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2} \: \: }}$

So, using this identity, we get

$\rm \: {(2x - a)}^{2} - {b}^{2} = 0$

We know,

$\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }}$

So, using this identity, we get

$\rm \: (2x - a - b)(2x - a + b) = 0$

$\rm \: 2x - a - b = 0 \: \: or \: \: 2x - a + b = 0$

$\rm \: 2x = a + b \: \: or \: \: 2x = a - b$

$\rm\implies \:x = \dfrac{a + b}{2} \: \: or \: \: x = \dfrac{a - b}{2}$

$\rule{190pt}{2pt}$

Alternative Method :-

$\rm \: {4x}^{2} - 4ax + {a}^{2} - {b}^{2} = 0$

can be rewritten as

$\rm \: {4x}^{2} - 2(2a)x + (a + b)(a - b) = 0$

$\rm \: {4x}^{2} - 2(a + a)x + (a + b)(a - b) = 0$

$\rm \: {4x}^{2} - 2(a + a + b - b)x + (a + b)(a - b) = 0$

$\rm \: {4x}^{2} - 2(a + b + a - b)x + (a + b)(a - b) = 0$

$\rm \: {4x}^{2} - 2(a + b)x - 2(a - b)x + (a + b)(a - b) = 0$

$\rm \: 2x(2x - a - b) - (a - b)(2x - a - b) = 0$

$\rm \: (2x - a - b)(2x - a + b) = 0$

$\rm \: 2x - a - b = 0 \: \: or \: \: 2x - a + b = 0$

$\rm \: 2x = a + b \: \: or \: \: 2x = a - b$

$\rm\implies \:x = \dfrac{a + b}{2} \: \: or \: \: x = \dfrac{a - b}{2}$

$\rule{190pt}{2pt}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

$\: \: \: \: \: \: \: \huge\color {pink}\boxed{\colorbox{black} {❥Answer}}$

The given quadratic equation is

4x² - 4ax + (a²-b²) = 0

4x² - 4ax + (a + b)(a - b) = 0

4x² +[- 2a-2a +2b-2b] x+(a - b) (a + b)=0

4x²+(2b-2a)x - (2a+2b)x +(a - b)(a + b)= 0

4x²+2(b - a)x - 2(a + b )x+ (a - b)(a + b)= 0

2x [2x - (a - b)]- (a + b) [2x - (a - b)] = 0

[2x - (a - b)] [2x - (a + b)] = 0

2x - (a - b) = 0, 2x - ( a + b ) = 0

$\sf \: ⇒x = \frac{a + b}{2} , \frac{a - b}{2}$