Question
Solve the following quadratic equation for $x : 4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
✓
Answer
$4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
This equation is of the form
$a x^2+b x+c=0$
Here $a=4, b=-4 a^2, c=\left(a^4-b^4\right)$
The quadratic formula to solve for $x$ is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{\left(-4 a^2\right)^2-4(4)\left(a^4-b^4\right)}}{2(4)}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{16 a^4-16 a^4+16 b^4}}{8}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{16 b^4}}{8} \Rightarrow x=\frac{4 a^2 \pm 4 b^2}{8}$
Dividing the numerator and denominator by $4$ we get,
$x=\frac{a^2 \pm b^2}{2}$
$\Rightarrow x=\frac{a^2+b^2}{2} \text { or } x=\frac{a^2-b^2}{2}$
Hence, the values of $x$ are $\frac{a^2+b^2}{2}$ and $\frac{a^2-b^2}{2}$.
This equation is of the form
$a x^2+b x+c=0$
Here $a=4, b=-4 a^2, c=\left(a^4-b^4\right)$
The quadratic formula to solve for $x$ is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{\left(-4 a^2\right)^2-4(4)\left(a^4-b^4\right)}}{2(4)}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{16 a^4-16 a^4+16 b^4}}{8}$
$\Rightarrow x=\frac{4 a^2 \pm \sqrt{16 b^4}}{8} \Rightarrow x=\frac{4 a^2 \pm 4 b^2}{8}$
Dividing the numerator and denominator by $4$ we get,
$x=\frac{a^2 \pm b^2}{2}$
$\Rightarrow x=\frac{a^2+b^2}{2} \text { or } x=\frac{a^2-b^2}{2}$
Hence, the values of $x$ are $\frac{a^2+b^2}{2}$ and $\frac{a^2-b^2}{2}$.
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