Question

Solve the equation : 4x2-4a2x+(a4-b4)=0

Answer 1

refer this file for answer

Answer 2

Answer to the equation: $\bold{\mathrm{x}=\frac{a^{2}-b^{2}}{2} \text { or } \mathrm{x}=\frac{a^{2}+b^{2}}}{2}$

To find:

Solve the equation: $4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$

Solution:

$4 \mathrm{x}^{2}-4 \mathrm{a}^{2} \mathrm{x}+\left(\mathrm{a}^{4}-\mathrm{b}^{4}\right)=0$

$\begin{array}{l}{(2 x)^{2}-\left[2\left(a^{2}+b^{2}\right)+2\left(a^{2}-b^{2}\right)\right] x+\left[\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)\right]=0} \\ \\{(2 x)^{2}-2 x\left(a^{2}+b^{2}\right)-2 x\left(a^{2}-b^{2}\right)+\left(a^{2}-b^{2}\right)+\left(a^{2}+b^{2}\right)=0}\end{array}$

$\begin{array}{l}{2 \mathrm{x}\left[2 \mathrm{x}-\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\right]-\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \times\left[2 \mathrm{x}-\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\right]=0} \\ \\ {\left[2 \mathrm{x}-\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)\right]\left[2 \mathrm{x}-\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\right]=0}\end{array}$

$\begin{array}{l}{2 x-\left(a^{2}-b^{2}\right)=0 \text { or } 2 x-\left(a^{2}+b^{2}\right)=0} \\ \\ {2 x=\left(a^{2}-b^{2}\right) \text { or } 2 x=\left(a^{2}+b^{2}\right)}\end{array}$

Hence, $\bold{\mathrm{x}=\frac{a^{2}-b^{2}}{2} \text { or } \mathrm{x}=\frac{a^{2}+b^{2}}{2}}$.