Question

solve:
${4x}^{2} + ( {a}^{2} - {b}^{2} ) = 0$
by b²-4ac method.....♪​

Answer 1

D=b^2-4ac=0-4(4)(a^2-b^2)
See the photo attached

Answer 2

solution:-

•The given equation is :- 4x²+(a²-b²)=0

we can write the equation:- 4x²+0x+(a²-b²)=0

comparing it with ax²+bx+c=0, we get

a=4,b=0 and c=(a²-b²)

find discriminant:- D=b²-4ac

D=b²-4ac

now put the value ,we get

D=(0)²-4×4×(a²-b²)

D= -16(a²-b²)

D= -16a²+16b²

D= 16b²-16a²

D=16(b²-a²)

Now use quadratic formula

$x = \frac{ - b \pm \sqrt{d} }{2a}$

put the value

$x = \frac{ - 0 \pm \sqrt{16( {b}^{2} - {a}^{2}) } }{2 \times 4}$

$x = \frac{ \pm \sqrt{16(b - a) {}^{2} } }{8}$

$x = \frac{ + 4(b - a)}{8} \: and \: x \frac{ - 4(b - a)}{8}$

$x = \frac{(b - a)}{2} \: and \: x \frac{ - (b - a)}{2}$

Answer:-

$\boxed { \large{x = \frac{(b - a)}{2} \: and \: x \frac{ - (b - a)}{2} }}$