Question

If $\alpha$ and $\beta$ are roots of the equation ${{x}^{2}}-px+q=0$ , then the quadratic equation whose roots are $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ is ___________.
(a) $q{{x}^{2}}-px+1=0$
(b) $q{{x}^{2}}+\left( {{p}^{2}}-2q \right)x+q=0$
(c) $q{{x}^{2}}+\left( 2q-{{p}^{2}} \right)x+q=0$
(d) $p{{x}^{2}}+\left( 2{{p}^{2}}-{{q}^{2}} \right)x+q=0$

Answer 1

First, we will find sum and products of roots of equation ${{x}^{2}}-px+q=0$ which is given as $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ respectively. Then, we are given that $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ are also the roots of equation so, we can write this as $\left( x-\dfrac{\alpha }{\beta } \right)\left( x-\dfrac{\beta }{\alpha } \right)=0$ . On multiplying these brackets and on simplification we will put the values of sum and product of roots which we already found. Thus, we will get the answer. Formula used here is ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .

Complete step-by-step solution:
Here, we are given that $\alpha$ and $\beta$ are roots of the equation ${{x}^{2}}-px+q=0$ . So, we can say that sum of the roots is $\dfrac{-b}{a}$ and product of roots is $\dfrac{c}{a}$ where $b=-p,c=q,a=1$ .
So, we can write it as
Sum of the roots $\alpha +\beta =\dfrac{-b}{a}=\dfrac{-\left( -p \right)}{1}$
$\alpha +\beta =p$ …………………………(1)
Product of roots $\alpha \beta =\dfrac{c}{a}=\dfrac{q}{1}=q$ …………………..(2)
Now, we have to find the equation whose roots are $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ . So, we can write this root as
$\left( x-\dfrac{\alpha }{\beta } \right)\left( x-\dfrac{\beta }{\alpha } \right)=0$
Now, we will multiply both the brackets. So, we get as
${{x}^{2}}-\dfrac{\beta x}{\alpha }-\dfrac{\alpha x}{\beta }+\dfrac{\alpha }{\beta }\cdot \dfrac{\beta }{\alpha }=0$
On further simplification, we can write this quadratic equation as
${{x}^{2}}-\left( \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \right)x+1=0$
We will take LCM, and further solving we get as
${{x}^{2}}-\left( \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)x+1=0$
Further we can write it as
$\alpha \beta {{x}^{2}}-\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)x+\alpha \beta =0$
Now, we will make $\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)$ perfect square equation by adding and subtracting $2\alpha \beta$ as we know the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ . So, we can write it as
$\alpha \beta {{x}^{2}}-\left( {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2\alpha \beta \right)x+\alpha \beta =0$
$\alpha \beta {{x}^{2}}-\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \right)x+\alpha \beta =0$
Now, we will put the values of equation (1) and (2) in the above equation. So, we get as
$q{{x}^{2}}-\left( {{\left( p \right)}^{2}}-2q \right)x+q=0$
Thus, we can write final equation as
$q{{x}^{2}}+\left( {{p}^{2}}-2q \right)x+q=0$
Hence, option (b) is the correct answer.

Note: Students might make mistakes by dividing the sum of roots with product of roots i.e. $\dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{p}{q}$ . On solving this we get as $\dfrac{\alpha }{\alpha \beta }+\dfrac{\beta }{\alpha \beta }=\dfrac{p}{q}$ . Thus, we get sum of roots of $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ as $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{p}{q}$ . Now, if we take the product of roots we will get as $\dfrac{\alpha }{\beta }\cdot \dfrac{\beta }{\alpha }=1$ . By taking this, we cannot directly take option value and find roots of such an equation. If we do this, then none of the options will directly match to the answer. So, this approach is wrong. Be careful in such types of problems.