Question

4x²-4ax+(a²-b²) = 0
solve the equation.

Answer 1

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

Given quadratic equation is

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

can be rewritten as

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

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

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

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

We know that,

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

So, using this identity, we get

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

We know,

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

Using this identity, we get

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

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

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

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

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