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# Ex.4.3 Q3 Quadratic Equations Solutions - NCERT Maths Class 10

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## Question

Find the roots of the following equations:

(i) \begin{align}x - \frac{1}{x} = 3,x \ne 0\end{align}

(ii)

\begin{align} & \frac{1}{{x + 4}} - \frac{1}{{x - 7}} = \frac{{11}}{{30}}, \\ \\ & x \ne - 4,\,7\end{align}

Video Solution
Ex 4.3 | Question 3

## Text Solution

What is known?

Quadratic equation, which is not in the form of $$ax^\text{2}+bx+c=0$$

What is Unknown?

Reasoning:

Convert the given equation in the form of \begin{align}ax^\text{2}+bx+c=0 \end{align} and by using the quadratic formula, find the roots.

Steps:

(i) \begin{align}x - \frac{1}{x} = 3,x \ne 0\end{align}

$$x - \frac{1}{x} = 3,x \ne 0$$ can be rewritten as (multiplying both sides by $$x$$):

\begin{align}{x^2} - 1 &= 3x\\{x^2} - 3x - 1 &= 0\end{align}

Comparing this against the standard form $$ax^\text{2}+bx+c=0$$, we find that:

$a = 1,\;b = - 3,\;c = - 1$

\begin{align}{{{b}}^2} - 4{{ac}}& = {( - 3)^2} - 4(1)( - 1)\\&= 9 + 4\\&= 13 > 0\end{align}

\begin{align}\therefore x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\&= \frac{{ - ( - 3) \pm \sqrt {13} }}{{2(1)}}\\x &= \frac{{3 \pm \sqrt {13} }}{2}\end{align}

The roots are \begin{align}\frac{{3 + \sqrt {13} }}{2},\frac{{3 - \sqrt {13} }}{2}.\end{align}

(ii) \begin{align}\frac{1}{{x + 4}} - \frac{1}{{x - 7}} = \frac{{11}}{{30}},\,x \ne - 4,\,7\end{align}

By cross multiplying we get:

\begin{align}\frac{{(x - 7) - (x + 4)}}{{(x + 4)(x - 7)}} &= \frac{{11}}{{30}}\\\frac{{x - 7 - x - 4}}{{{x^2} + 4x - 7x - 28}} &= \frac{{11}}{{30}} \\ \frac{{ - 11}}{{{x^2} - 3x - 28}} &= \frac{{11}}{{30}}\end{align} \begin{align} - 11 \times 30\! &\!= \!\!11 \! \left({x^2} - 3x - 28 \right) \\- 30 &= {x^2} - 3x - 28 \end{align} \begin{align} {x^2} - 3x - 28 + 30 &= 0\\{x^2} - 3x + 2 &= 0\end{align}

Comparing this against the standard form $$ax^\text{2}+bx+c=0$$, we find that:

$a = 1\;b = - 3,\;c = 2$

\begin{align}{{b^2} - 4ac}& = {{( - 3)}^2} - 4(1)(2)\\&= 9 - 8\\&= 1 > 0\end{align}

$$\therefore \;$$Real roots exist for this quadratic equation.

\begin{align}x&=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\{x}&=\frac{-(-3)\pm \sqrt{{{(-3)}^{2}}-4(1)(2)}}{2(1)} \\{x}&=\frac{3\pm 1}{2} \\{x} &=\frac{3+1}{2}\quad {x}=\frac{3-1}{2} \\ x &=\frac{~4~}{2} \qquad \;x=\frac{2}{2} \\ x&=2 \qquad \;\;\;\;x=1 \\\end{align}

Roots are $$2, \,1.$$

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