Question

Solve the quadratic equations by the method of completing the square:4x² +4bx-(a²-b²)=0

Answer 1

Answer:

Now in the above quadratic equation the coefficient of x2 is 4. Let us make it unity by dividing the entire quadratic equation by 4.

4x2 + 4bx – (a2 – b2) = 0

x2 + bx = (a2 – b2)/4

Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.

Coefficient of x = b

Half of b = b/2

Squaring the half of b = b/4

Now the LHS term is a perfect square and can be expressed in the form of (a-b) 2 = a2 – 2ab + b2 where a = x and b = b/2

On simplifying both RHS and LHS we get an equation of following form,

(x ± A)2 = k2

Taking Square root of both sides.

Now taking the positive part,

x = (a – b) / 2

Now taking the negative part,

x = - (a + b) / 2

Answer 2

SOLUTION

We write, 4bx = 2(a+b)x -2(a-b)x as

$= > 4 {x}^{2} \times [ - ( {a}^{2} - {b}^{2} )] = - 4( {a}^{2} - {b}^{2}) {x}^{2} \\ \\ = > 2(a + b)x \times [ - 2(a - b)x] \\ \\ 4 {x}^{2} + 4bx - (a {}^{2} - {b}^{2} ) = 0 \\ \\ = > 4 {x}^{2} + 2(a + b)x - 2(a - b)x - (a - b)(a + b) = 0 \\ \\ = > 2x[2x + (a + b)]- (a - b)[2x + (a + b)] = 0 \\ \\ = > [2x + (a + b)][2x - (a - b)] = 0 \\ \\ = > 2x + (a + b) = 0 \: \: or \: \: 2x - (a - b) = 0 \\ \\ = > x = - \frac{a + b}{2} \: \: or \: \: x = \frac{a - b}{2}$

Hope it helps ☺️