Login/Sign Up
Home » Find the roots of the each of the following equations, if they exist, by applying the quadratic formula:
Find the roots of the each of the following equations, if they exist, by applying the quadratic formula: \frac{1}{x}-\frac{1}{x-2}=3, x \neq 0,2
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{1}{x}-\frac{1}{x-2}=3, x \neq 0,2

The given equation is

\begin{array}{l} \frac{1}{x}-\frac{1}{x-2}=3, x \neq 0,2 \\ \Rightarrow \frac{x-2-x}{x(x-2)}=3 \\ \Rightarrow \frac{-2}{x^{2}-2 x}=3 \\ \Rightarrow-2=3 x^{2}-6 x \\ \Rightarrow 3 x^{2}-6 x+2=0 \end{array}

This equation is of the form a x^{2}+b x+c=0, where a=3, b=-6 and c=2.

\therefore Discriminant, D=b^{2}-4 a c=(-6)^{2}-4 \times 3 \times 2=36-24=12>0

So, the given equation has real roots.

Now, \sqrt{D}=\sqrt{12}=2 \sqrt{3}

\begin{array}{l} \therefore a=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-6)+2 \sqrt{3}}{2 \times 3}=\frac{6+2 \sqrt{3}}{6}=\frac{3+\sqrt{3}}{3} \\ \beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-6)-2 \sqrt{3}}{2 \times 3}=\frac{6-2 \sqrt{3}}{6}=\frac{3-\sqrt{3}}{3} \end{array}

Hence, \frac{3+\sqrt{3}}{3} and \frac{3-\sqrt{3}}{3} are the roots of the given equation.

i

More Material