Question

The value of $a$ for which one root of the equation $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ is twice as large as other
(A) $\dfrac{-1}{3}$
(B) $\dfrac{2}{3}$
(C) $\dfrac{-2}{3}$
(D) $\dfrac{1}{3}$

Answer 1

Assume that $\alpha$ and $\beta$ are the two roots of the quadratic equation $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ such that $\beta$ is twice as large as $\alpha$ . So, $2\alpha =\beta$ . Compare the quadratic equation $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ with the standard form of the quadratic equation, $p{{x}^{2}}+qx+r=0$ and get the values of p, q, and r. We know the formula, the sum of all roots is equal to $\dfrac{-q}{p}$ and the product of all roots is equal to $\dfrac{r}{p}$ . Use these formulas and get the value of $\left( \alpha +\beta \right)$ and $\alpha \beta$ . Now, replace $\beta$ by $2\alpha$ . Here, we have two equations in $\alpha$ and $a$ . Now, solve it further and get the value of a.

Complete step-by-step solution:
According to the question, it is given that
The quadratic equation = $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ ……………………………………………….(1)
It is also given that the one root is twice as large as the other root.
First of all, let us assume that $\alpha$ and $\beta$ are the two roots of the quadratic equation $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ such that $\beta$ is twice as large as $\alpha$ .
So, we can say that,
$2\alpha =\beta$ …………………………………………(2)
We know the standard form of the quadratic equation, $p{{x}^{2}}+qx+r=0$ .
From equation (1), we have the quadratic equation.
Now, on comparing equation the quadratic equation $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ and $p{{x}^{2}}+qx+r=0$ , we get
$p=\left( {{a}^{2}}-5a+3 \right)$ ……………………………………(3)
$q=\left( 3a-1 \right)$ ………………………………………..(4)
$r=2$ ……………………………………………(5)
We know the formula for the quadratic equation $p{{x}^{2}}+qx+r=0$ ,
The sum of all roots = $\dfrac{-q}{p}$ …………………………………..(6)
The product of all roots = $\dfrac{r}{p}$ ………………………………………….(7)
For the quadratic equation $\left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0$ , we have $\alpha$ and $\beta$ as its roots.
Now, from equation (3), equation (4), and equation (6), we get
The sum of all roots = $~\alpha +\beta =\dfrac{-\left( 3a-1 \right)}{\left( {{a}^{2}}-5a+3 \right)}$ …………………………………..(8)
Now, from equation (3), equation (5), and equation (7), we get
The product of all roots = $\alpha \times \beta =\dfrac{2}{\left( {{a}^{2}}-5a+3 \right)}$ ……………………………………….(9)
Now, from equation (2) and equation (8), we get

& \Rightarrow ~\alpha +2\alpha =\dfrac{-\left( 3a-1 \right)}{\left( {{a}^{2}}-5a+3 \right)} \\\ & \Rightarrow 3\alpha =\dfrac{-\left( 3a-1 \right)}{\left( {{a}^{2}}-5a+3 \right)} \\\ \end{aligned}$$ $$\Rightarrow \alpha =\dfrac{-\left( 3a-1 \right)}{3\left( {{a}^{2}}-5a+3\right)}$$……………….(10) Now, from equation (2) and equation (9), we get $$\begin{aligned} & \Rightarrow \alpha \times \left( 2\alpha \right)=\dfrac{2}{\left( {{a}^{2}}-5a+3 \right)} \\\ & \Rightarrow 2{{\alpha }^{2}}=\dfrac{2}{\left( {{a}^{2}}-5a+3 \right)} \\\ \end{aligned}$$ $$\Rightarrow {{\alpha }^{2}}=\dfrac{1}{\left( {{a}^{2}}-5a+3 \right)}$$ ……………………(11) Now, on putting the value of $$\alpha $$ from equation (10) in equation (11), we get $$\begin{aligned} & \Rightarrow {{\left\\{ \dfrac{-\left( 3a-1 \right)}{3\left( {{a}^{2}}-5a+3 \right)} \right\\}}^{2}}=\dfrac{1}{\left( {{a}^{2}}-5a+3 \right)} \\\ & \Rightarrow \dfrac{{{\left( 1-3a \right)}^{2}}}{9{{\left( {{a}^{2}}-5a+3 \right)}^{2}}}=\dfrac{1}{\left( {{a}^{2}}-5a+3 \right)} \\\ & \Rightarrow {{\left( 1-3a \right)}^{2}}=9\left( {{a}^{2}}-5a+3 \right) \\\ & \Rightarrow \left( 1+9{{a}^{2}}-6a \right)=9{{a}^{2}}-45a+27 \\\ & \Rightarrow 45a-6a=27-1 \\\ & \Rightarrow 39a=26 \\\ & \Rightarrow a=\dfrac{26}{39} \\\ & \Rightarrow a=\dfrac{2}{3} \\\ \end{aligned}$$ Therefore, the value of $$a$$ is $$\dfrac{2}{3}$$ . **Hence, the correct option is (B).** **Note:** To solve this question, one might try to solve using the perfect square method. But this approach will be wrong because applying the perfect square method to get the roots of the quadratic equation $$\left( \left( {{a}^{2}}-5a+3 \right){{x}^{2}}+\left( 3a-1 \right)x+2=0a \right)$$ is not an easy task here. So, don’t approach this question by using the perfect square method.