Question

The roots of the equation ${{x}^{2}}+ax+b=0$are p and q, then the equation whose roots are ${{p}^{2}}q$ and $p{{q}^{2}}$will be
( a ) ${{x}^{2}}+abx+{{b}^{3}}=0$
( b ) ${{x}^{2}}-abx+{{b}^{3}}=0$
( c ) $b{{x}^{2}}+x+a=0$
( d ) ${{x}^{2}}+ax+ab=0$

Answer 1

Hint: We are given a quadratic equation with its roots and we have to find the equation whose roots are ${{p}^{2}}q$and $p{{q}^{2}}$. By finding out the sum and the product of roots, we are able to find out the equation and find the correct answer.

Formula Used: If $\alpha $ and $\beta $are the roots the in the standard form of quadratic equation $a{{x}^{2}}+bx+c=0$
Then sum of roots $(\alpha +\beta )=-\dfrac{b}{a}$ and the product of roots $(\alpha \beta )=\dfrac{c}{a}$

Complete step by step Solution:
Given the quadratic equation ${{x}^{2}}+ax+b=0$…………………………….. ( 1)
And the roots of the equation are p and q.
We have to find the equation whose roots are ${{p}^{2}}q$and $p{{q}^{2}}$
We know if $\alpha $ and $\beta $are the roots the in the standard form of quadratic equation $a{{x}^{2}}+bx+c=0$
Then sum of roots $(\alpha +\beta )=-\dfrac{b}{a}$ and the product of roots $(\alpha \beta )=\dfrac{c}{a}$
So the sum of roots ( p + q) = - a and the product of roots (pq) = b
Now the required equation whose roots are ${{p}^{2}}q$and $p{{q}^{2}}$
Therefore the sum of roots = ${{p}^{2}}q+p{{q}^{2}}$
That is pq ( p + q ) = -ab
And the product of roots = ${{p}^{2}}q\times p{{q}^{2}}$
Which is equal to ${{(pq)}^{3}}$ = ${{b}^{3}}$
Thus the equation is ${{x}^{2}}+abx+{{b}^{3}}=0$

Therefore, the correct option is (a).

Note: Whenever the sum and the product of the roots are given in the question, it is the easiest way to solve the question with the formula
Sum of roots = $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$
And the product of roots = $\dfrac{ constant\, term}{coefficient\, of\, x^2}$
With this, we are able to solve the question fastly and accurately.