Question
Solve the following quadratic equations for $x :4 x^2+4 b x-\left(a^2-b^2\right)=0$
✓
Answer
The given quadratic equation is
$4 x^2+4 b x-\left(a^2-b^2\right)=0$
So, $a=4, b=4 b, c=-\left(a^2-b^2\right)$
Quadratic formula to find the roots is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$x=\frac{-4 b \pm \sqrt{(4 b)^2-4 \times 4 \times\left\{-\left(a^2-b^2\right)\right\}}}{2 \times 4}$
$x=\frac{-4 b \pm \sqrt{16 b^2+16\left(a^2-b^2\right)}}{2 \times 4}$
$x=\frac{-4 b \pm \sqrt{16 b^2+16 a^2-16 b^2}}{8}$
$x=\frac{-4 b \pm \sqrt{16 a^2}}{8}$
$x=\frac{-4 b \pm 4 a}{8}$
$x=\frac{-b \pm a}{2}$
$x=\frac{-b+a}{2}, \frac{-b-a}{2}$
$4 x^2+4 b x-\left(a^2-b^2\right)=0$
So, $a=4, b=4 b, c=-\left(a^2-b^2\right)$
Quadratic formula to find the roots is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$x=\frac{-4 b \pm \sqrt{(4 b)^2-4 \times 4 \times\left\{-\left(a^2-b^2\right)\right\}}}{2 \times 4}$
$x=\frac{-4 b \pm \sqrt{16 b^2+16\left(a^2-b^2\right)}}{2 \times 4}$
$x=\frac{-4 b \pm \sqrt{16 b^2+16 a^2-16 b^2}}{8}$
$x=\frac{-4 b \pm \sqrt{16 a^2}}{8}$
$x=\frac{-4 b \pm 4 a}{8}$
$x=\frac{-b \pm a}{2}$
$x=\frac{-b+a}{2}, \frac{-b-a}{2}$
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