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
Quadratic Equations
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?

Roots of a quadratic equation.

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:

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

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 \(2, \,1.\)