Question

If p and q are the zeroes of the quadratic polynomial 2x^2 + 2(m+n)x + m^2 + n^2 , form the quadratic polynomial whose zeroes are (p+q)^2 and (p-q)^2

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Answer 1

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Sum of zeroes/roots p+q=-m____(1)

Product of zeroes p*q=n^2____(2)

Squaring equation (1) on both sides,

p^2 + q^2 + 2pq = m^2_____(3)

Substituting value of p*q from equation (2) in (3)

p^2 + q^2 + 2n^2 =m^2

Therefore, p^2 + q^2 =m^2 - 2n^2

So value of p^2 + q^2 +pq

=m^2 - 2n^2 +n^2 (since pq=n^2 from (2) )

=m^2 - n^2 or

=(m+n)(m-n)

Answer 2

Use Vieta's method on $2x^2 + 2(m+n)x + m^2 + n^2$.

$\displaystyle\left \{ {{p + q=-(m + n)} \atop {pq =\dfrac{m^2 + n^2}{2}}} \right.$

Two zeros of the new polynomial:

$(p+q)^2=p^2+2pq+q^2$ and $(p-q)^2=p^2-2pq+q^2$

Construct the new polynomial with Vieta's method.

Sum and product of the new polynomial:

Sum $2(p^2+q^2)$

Product $(p+q)^2(p-q)^2$

Finding the sum:

$(p+q)^2-2pq=p^2+q^2$

$=(m+n)^2-(m^2+n^2)$

$=2mn$

→ $4mn$ is the sum.

Finding the product:

$(p+q)^2-4pq=(p-q)^2$

$=(m+n)^2-2(m^2+n^2)$

$=-(m^2-2mn+n^2)=-(m-n)^2$

→  $-(m+n)^2(m-n)^2$ is the product.

The new quadratic equation is $x^2-4mnx-(m+n)^2(m-n)^2$.

More information:

Vieta's Method

Consider a quadratic polynomial $x^2+\dfrac{b}{a} x+\dfrac{c}{a}$.

If α and β are the zeroes of the polynomial then

$(x-\alpha )(x-\beta )=x^2-(\alpha +\beta )x+\alpha \beta\;\textbf{[Factor Theorem]}$.

$\alpha +\beta$ is the sum of the two zeroes.

$\alpha \beta$ is the product of the two zeroes.

So $\alpha +\beta =-\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}$.