Question

If $\alpha ,\beta$ are the roots of the equation $x^2-px+q=0$, then the quadratic equation whose roots are $({\alpha }^2-{\beta }^2)({\alpha }^3-{\beta }^3)$ and $({\alpha }^3{\beta }^2+{\alpha }^2{\beta }^3)$:

• A. $q{x}^2-(5q+7p)x-(q^6{p}^6+4q^2{p}^6)=0$
• B. $p{x}^2-(5p+7q)x-(p^6{q}^6+4p^2{q}^6)=0$
• C. $x^2-(p^4-5p^{3}q+5p{q}^2)x+(p^6{q}^2-5p^4{q}^3+4p^2{q}^4)=0$
• D. $x^2-(p^{3}q-5p^5+p^{4}q)x-(p^6{q}^2-5p^2{q}^6)=0$

Answer 1

The correct option is C $x^2-(p^4-5p^{3}q+5p{q}^2)x+(p^6{q}^2-5p^4{q}^3+4p^2{q}^4)=0$

When this problem will be solved by algebraic methods, it will take too much time to solve beyond the normal required times. So, the best way to get the correct and quick answer is to assume some simple roots (Le., $\alpha$ and $\beta$) then go through options.
Let us take two arbitrary values $\alpha =-1,\beta =2$, then the equation will be $x^2-x-2=0$
Comparing with the equation $x^2-px+q=0$
$\Rightarrow p=1,q=-2$
Now, the sum of the roots of the required equation
$=\left[\right({\alpha }^2-{\beta }^2\left)\right({\alpha }^3-{\beta }^3\left)\right]+[{\alpha }^3{\beta }^2+{\alpha }^2{\beta }^3]=27+4=31$
and product of roots $[{\alpha }^2-{\beta }^2][{\alpha }^3-{\beta }^3][{\alpha }^3{\beta }^2+{\alpha }^2{\beta }^3]=27\times 4=108$
Hence equation is $x^2-31x+108=0$
Now putting the values of p and q in the equation options a, b, and c we get option (c) is correct. as:
$x^2-[1-\{5\times 1(-2\left)\right\}+5\times 1\times 4]+[1\times 4-\{51\times (-8\left)\right\}+4\times 1\times 16]=0=x^2-31x+108=0$