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# Find the roots of following equations:(1) $x-\dfrac{1}{x}=3,x\ne 0$(2) $\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne -4,7$

Last updated date: 09th Aug 2024
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Hint: First we have to find the value of discriminant ${{b}^{2}}-4ac$ and check if the roots are real and unequal or not. Then substituting the values of a, b, c in formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the roots of the equation.

(1) We know the general quadratic equation is given as $a{{x}^{2}}+bx+c=0$ where a, b are the coefficients of ${{x}^{2}}$ and x respectively and c is the constant in the equation.
By expressing the given equation $x-\dfrac{1}{x}=3$ in the form of a general quadratic equation we get,
Multiplying with x on both sides of the equation,
${{x}^{2}}-1=3x$
By subtracting with 3x on both sides we get,
${{x}^{2}}-3x-1=0$
Comparing the above equation with the general quadratic equation we get the values of a, b, c as,
$a=1$, $b=-3$, $c=-1$
We know the formula to find the roots of a general quadratic equation is given as $\dfrac{-b\pm \sqrt{D}}{2a}$ where $D={{b}^{2}}-4ac$ is known as the discriminant.
Applying the above formula by substituting the values of a, b, c to find the discriminant,
For the roots to be real and unequal the discriminant must be greater than 0.
$\left[ {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right) \right]>0$
Squaring $-3$ and multiplying $-4$ with$-1$ we get,
$\left( 9+4 \right)>0$
Adding 9 with 4 the resultant is greater than 0,
13 > 0
Hence the roots are real and unequal.
Applying the formula to find the roots of a general quadratic equation by substituting the values of a, b, c and the value of the discriminant ${{b}^{2}}-4ac$ we get,
$\dfrac{-\left( -3 \right)\pm \sqrt{13}}{2}$
By taking square root of 13 and simplifying the expression for $\dfrac{-b+\sqrt{D}}{2a}$ and $\dfrac{-b-\sqrt{D}}{2a}$ we get,
\begin{align} & x=3.3027 \\ & x=-0.3027 \\ \end{align}
Hence the roots of the equation $x-\dfrac{1}{x}=3$ are 3.3027 and -0.3027.
(2)
By expressing the given equation $\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}$ in the form of a general quadratic equation we get,
Taking a common denominator $\left( x+4 \right)\left( x-7 \right)$ in the LHS,
$\dfrac{\left( x-7 \right)-\left( x+4 \right)}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}$
Simplifying the above equation,
$\dfrac{x-7-x-4}{\left( x+4 \right)\left( x-7 \right)}=\dfrac{11}{30}$
$\dfrac{-11}{{{x}^{2}}-7x+4x-28}=\dfrac{11}{30}$
Cross multiplying the terms to express as a quadratic equation,
Cancelling out 11 and multiplying with $-1$ on both sides,
${{x}^{2}}-3x-28=-30$
Expressing the equation in the simplest form possible,
${{x}^{2}}-3x+2=0$
Comparing the above equation with the general quadratic equation we get the values of a, b, c as,
$a=1$, $b=-3$, $c=2$
We know the formula to find the roots of a general quadratic equation is given as $\dfrac{-b\pm \sqrt{D}}{2a}$ where $D={{b}^{2}}-4ac$ is known as the discriminant.
Applying the above formula by substituting the values of a, b, c to find the discriminant,
For the roots to be real and unequal the discriminant must be greater than 0.
$\left[ {{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( 2 \right) \right]>0$
Squaring $-3$ and multiplying $-4$ with 2 we get,
$\left( 9-8 \right)>0$
Subtracting 8 from 9 the resultant is greater than 0,
1 > 0
Hence the roots are real and unequal.
Applying the formula to find the roots of a general quadratic equation by substituting the values of a, b, c and the value of the discriminant ${{b}^{2}}-4ac$ we get,
$\dfrac{-\left( -3 \right)\pm \sqrt{1}}{2}$
By taking square root of 1 and simplifying the expression for $\dfrac{-b+\sqrt{D}}{2a}$ and $\dfrac{-b-\sqrt{D}}{2a}$ we get,
x = 2
x = 1
Hence the roots of the equation $\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}$ are 2 and 1.

So, the correct answer is “Option A”.

Note: The discriminant plays a major role which specifies the nature of roots therefore calculate the discriminant D carefully. The roots can be either real or imaginary which are determined by discriminant. If the value of discriminant is less than 0 then no real roots exist and the roots are imaginary.