1. Find the root of the quadratic equation $$bx^{2}-2ax+a=0$$





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  • By: anil on 05 May 2019 02.40 am
    If the quadratic equation is of type $$Ax^+Bx+C=0$$ then the roots of the quadratic equation are given by $$x = dfrac{-Bpmsqrt{B^2 - 4AC}}{2A}$$ Comparing $$Ax^+Bx+C=0$$ with $$bx^{2}-2ax+a=0$$, A = b,  B = -2a,  C = a Hence, the roots = $$dfrac{2apmsqrt{4a^2 - 4ba}}{2b}$$ x = $$dfrac{apmsqrt{a^2 - ba}}{b}$$
    Let $$x_{1}$$ = $$dfrac{a-sqrt{a^2 - ba}}{b}$$,  $$x_{2}$$ = $$dfrac{a+sqrt{a^2 - ba}}{b}$$ Rationalizing  $$x_{1}$$ = $$dfrac{a-sqrt{a^2 - ba}}{b}$$ $$Rightarrow$$ $$x_{1}$$ = $$dfrac{a-sqrt{a^2 - ba}}{b}*dfrac{a+sqrt{a^2 - ba}}{a+sqrt{a^2 - ba}}$$ $$Rightarrow$$ $$x_{1}$$ = $$dfrac{a^2-(a^2 - ba)}{b*(a+sqrt{a^2 - ba})}$$
    $$Rightarrow$$ $$x_{1}$$ = $$dfrac{ab}{b*(a+sqrt{a^2 - ba})}$$
    $$Rightarrow$$ $$x_{1}$$ = $$dfrac{a}{a+sqrt{a^2 - ba}}$$
    $$Rightarrow$$ $$x_{1}$$ = $$dfrac{a}{a+sqrt{a}*sqrt{a - b}}$$
    $$Rightarrow$$ $$x_{1}$$ = $$dfrac{sqrt{a}}{sqrt{a}+sqrt{a - b}}$$
    Similarly, $$x_{2}$$ = $$dfrac{sqrt{a}}{sqrt{a}-sqrt{a - b}}$$ Therefore, x = $$dfrac{sqrt{a}}{sqrt{a}pmsqrt{a - b}}$$. Hence, option C is the correct answer. 
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