Question

10. Find quadratic equation such that its roots are square of sum of the roots
square of difference of the roots of equation 2x+ + 2(p + q) x + p2 + q2 = 0​

Answer 1

Let's assume roots are [math]m[/math] and [math]n[/math].

So, we want the equation whose roots would be [math](m+n)^2 and (m-n)^2[/math]

So, the sum of the roots(S) of our desired equation would be [math]2(m^2+n^2) [/math]and product of the roots(P) would be [math](m+n)^2(m-n)^2[/math].

What we know from given equation are:

[math]m+n=-(p+q)[/math]

And

[math]mn= (p^2+q^2)/2[/math]

The Sum and Product are :

[math]S=2(m^2+n^2)=2{(m+n)^2-2mn}[/math]

[math]=2{(p+q)^2-(p^2+q^2)}=2*2pq=4pq[/math]

And

[math]P=(m+n)^2(m-n)^2=(p+q)^2{(m+n)^2-4mn}[/math]

[math]=(p+q)^2{(p+q)^2-2(p^2+q^2}[/math]

[math]=(p+q)^2(2pq-p^2-q^2)[/math]

[math]=-(p+q)^2(p-q)^2[/math]

[math]=-(p^2-q^2)^2[/math]

Our desired equation would be [math]x^2-Sx+P=0[/math]

So, [math]x^2-4pqx-(p^2-q^2)^2=0[/math] is our required equation.

I hope this may help you

Answer 2

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