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Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: \frac{m}{n} x^{2} \frac{n}{m}=1-2 x
CBSE | Class 10 | Exercise 10C | Maths | Quadratic Equations | RS Aggarwal
Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: \frac{m}{n} x^{2} \frac{n}{m}=1-2 x

The given equation is

\begin{array}{l} \frac{m}{n} x^{2} \frac{n}{m}=1-2 x \\ \Rightarrow \frac{m^{2} x^{2}+n^{2}}{m n}=1-2 x \\ \Rightarrow m^{2} x^{2}+n^{2}=m n-2 m n x \\ \Rightarrow m^{2} x^{2}+2 m n x+n^{2}-m n=0 \end{array}

This equation is of the form a x^{2}+b x+c=0, where a=m^{2}, b=2 m n and c=n^{2}-m n \therefore Discriminant,

D=b^{2}-4 a c=(2 m n)^{2}-4 \times m^{2} \times\left(n^{2}-m n\right)=4 m^{2} n^{2}-4 m^{2} n^{2}+4 m^{3} n^{2}=4 m^{3} n>0

So, the given equation has real roots.

Now, \sqrt{D}=\sqrt{4 m^{3} n}=2 m \sqrt{m n}

\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-2 m n+2 m \sqrt{m n}}{2 \times m^{2}}=\frac{2 m n(-n+\sqrt{m n})}{2 m^{2}}=\frac{-n+\sqrt{m n}}{m}

\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-2 m n-2 m \sqrt{m n}}{2 \times m^{2}}=\frac{-2 m(n+\sqrt{m n})}{2 m^{2}}=\frac{-n+\sqrt{m n}}{m}

Hence, \frac{-n+\sqrt{m n}}{m} and \frac{-n-\sqrt{m n}}{m} are the roots of the given equation.

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