question_answer

Solve the following quadratic equation for x:   \[4{{x}^{2}}+4bx-\left( {{a}^{2}}-{{b}^{2}} \right)=0\]

Answer:

The given equation is

\[4{{x}^{2}}+4bx-\left( {{a}^{2}}-{{b}^{2}} \right)=0\]                           ...(i)

Comparing equation (i) with quadratic equation

\[A{{x}^{2}}+Bx+C=0\], we get

\[A=4,\text{ }B=4b,\text{ }C=-\left( {{a}^{2}}-{{b}^{2}} \right)\]

By quadratic formula

\[x=\frac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}\]

\[x=\frac{-4b\pm \sqrt{16{{b}^{2}}+4\times 4\times ({{a}^{2}}-{{b}^{2}})}}{2\times 4}\]

\[x=\frac{-4b\pm \sqrt{16{{b}^{2}}+16{{a}^{2}}-16{{b}^{2}})}}{8}\]

\[x=\frac{-4b\pm 4a}{8}\]

\[x=\frac{-b\pm a}{2}\]

Therefore,          \[x=\frac{-b-a}{2}\Rightarrow -\left( \frac{a+b}{2} \right)\]

or                     \[x=\frac{-b+a}{2}\Rightarrow \frac{a-b}{2}\]

Hence,  \[x=-\left( \frac{a+b}{2} \right)\] and \[x=\frac{a-b}{2}\].