Find the root of the quadratic equation bx²-2ax+a=0

Correct Answer is : aa±a−b{{a}}{{a}±{a-b}}a±a−ba

Correct Answer is : aa±a−b{{a}}{{a}±{a-b}}a±a−ba

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Indian Institute of Foreign Trade 2010 Solved Paper

Question : 11 of 135

Marks: +1, -0

bx^{2}-2ax+a=0

Solution:

If the quadratic equation is of type

Ax+Bx+C=0

then the roots of the quadratic equation are given by

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

Comparing

Ax+Bx+C=0

with

bx^{2}-2ax+a=0

A=b,B=-2a,C=a

Hence, the roots =

\frac{2a\pm\sqrt{4a^{2}-4ba}}{2b}

x=\frac{a\pm\sqrt{a^{2}-ba}}{b}

Let

x_{1}=\frac{a-\sqrt{a^{2}-ba}}{b},x_{2}=\frac{a+\sqrt{a^{2}-ba}}{b}

Rationalizing

x_{1}=\frac{a-\sqrt{a^{2}-ba}}{b}

\Rightarrow x_{1}=\frac{a-\sqrt{a^{2}-ba}}{b}\times\frac{a+\sqrt{a^{2}-ba}}{a+\sqrt{a^{2}-ba}}

\Rightarrow x_{1}=\frac{a^{2}-(a^{2}-ba)}{b\times(a+\sqrt{a^{2}-ba})}

\Rightarrow x_{1}=b\times\frac{ab}{a+\sqrt{a^{2}-ba}}

\Rightarrow x_{1}=a+\frac{a}{\sqrt{a^{2}-ba}}

\Rightarrow x_{1}=a+\frac{a}{\sqrt{a}\times\sqrt{a-b}}

\Rightarrow x_{1}=a+\frac{a}{\sqrt{a}+\sqrt{a-b}}

Similarly,

x_{2}=\frac{\sqrt{a}}{\sqrt{a}-\sqrt{a-b}}

Therefore,

x=\frac{\sqrt{a}}{\sqrt{a}\pm\sqrt{a-b}}

Hence, option C is the correct answer.

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