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