Question

Find the roots of the quadratic equation $\dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}$, where $x\ne 3,5$.

Answer 1

Hint: We start solving the problem by adding the two fractions containing the terms of x into a single fraction by taking the LCM. We then multiply the terms present on both sides (i.e., multiplying the numerator of L.H.S with the denominator of R.H.S and numerator of R.H.S with the denominator of L.H.S). We then factorize the obtained resultant quadratic equation to get the required values of the roots of the quadratic equation.

Complete step by step answer:
According to the problem, we need to find the roots of the quadratic equation $\dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}$, where $x\ne 3,5$.
Let us convert the given quadratic equation $\dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}$, where $x\ne 3,5$ into its simpler form.
$\Rightarrow \dfrac{\left( x-2 \right)\left( x-5 \right)+\left( x-3 \right)\left( x-4 \right)}{\left( x-3 \right)\left( x-5 \right)}=\dfrac{10}{3}$.
$\Rightarrow \dfrac{{{x}^{2}}-7x+10+{{x}^{2}}-7x+12}{{{x}^{2}}-8x+15}=\dfrac{10}{3}$.
$\Rightarrow \dfrac{2{{x}^{2}}-14x+22}{{{x}^{2}}-8x+15}=\dfrac{10}{3}$.
Dividing by 2 on both sides, we get
$\Rightarrow \dfrac{{{x}^{2}}-7x+11}{{{x}^{2}}-8x+15}=\dfrac{5}{3}$.
Let us cross multiply the terms present on both sides.
$\Rightarrow 3\times \left( {{x}^{2}}-7x+11 \right)=5\times \left( {{x}^{2}}-8x+15 \right)$.
$\Rightarrow 3{{x}^{2}}-21x+33=5{{x}^{2}}-40x+75$.
$\Rightarrow 2{{x}^{2}}-19x+42=0$.
Let us factorize the obtained quadratic equation.
$\Rightarrow 2{{x}^{2}}-12x-7x+42=0$.
$\Rightarrow 2x\left( x-6 \right)-7\left( x-6 \right)=0$.
$\Rightarrow \left( 2x-7 \right)\left( x-6 \right)=0$.
$\Rightarrow \left( 2x-7 \right)=0$ or $\left( x-6 \right)=0$.
$\Rightarrow 2x=7$ or $x=6$.
$\Rightarrow x=\dfrac{7}{2}$ or $x=6$.
So, we found the roots of the given quadratic equation as $\dfrac{7}{2}$ and 6.
∴ The roots of the quadratic equation $\dfrac{x-2}{x-3}+\dfrac{x-4}{x-5}=\dfrac{10}{3}$, where $x\ne 3,5$ are $\dfrac{7}{2}$ and 6.

Note: Whenever we get this type of problem containing quadratic equations involved with fractions of the independent variable, we first need to convert it to the simpler form and check the domain of it. We can also solve the resultant quadratic equation $2{{x}^{2}}-19x+42=0$ by using the facts that roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
i.e., the roots of quadratic equation $2{{x}^{2}}-19x+42=0$ are $\dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}$.
$\Rightarrow \dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}=\dfrac{19\pm \sqrt{361-336}}{4}$.
$\Rightarrow \dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}=\dfrac{19\pm \sqrt{25}}{4}$.
$\Rightarrow \dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}=\dfrac{19\pm 5}{4}$.
$\Rightarrow \dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}=\dfrac{19+5}{4},\dfrac{19-5}{4}$.
$\Rightarrow \dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}=\dfrac{24}{4},\dfrac{14}{4}$.
$\Rightarrow \dfrac{-\left( -19 \right)\pm \sqrt{{{\left( -19 \right)}^{2}}-4\left( 2 \right)\left( 42 \right)}}{2\left( 2 \right)}=6,\dfrac{7}{2}$, which are the required roots of the given quadratic equation.