Multiple choice

Find the equation whose roots are $\displaystyle \left ( \alpha +\beta \right )^{2}:and:\left ( \alpha -\beta \right )^{2},$ where $\displaystyle \alpha$ and $\beta $ are the roots of $\displaystyle 2x^{2}+2\left ( m+n \right )x+\left ( m^{2}+n^{2} \right )= 0.$

  1. $\displaystyle x^{2}-4mnx-\left ( m^{2}+n^{2} \right )^{2}= 0.$
  2. $\displaystyle x^{2}+2mnx-\left ( m^{2}-n^{2} \right )^{2}= 0.$
  3. $\displaystyle x^{2}-4mnx-\left ( m^{2}-n^{2} \right )^{2}= 0.$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
AI explanation

Let the new roots be R1 = (alpha + beta)^2 and R2 = (alpha - beta)^2, and for the original equation 2x^2 + 2(m+n)x + (m^2 + n^2) = 0, the sum and product of the roots are alpha + beta = -(m+n) and alpha*beta = (m^2 + n^2)/2. The sum of the new roots is R1 + R2 = 2(alpha^2 + beta^2) = 2((alpha + beta)^2 - 2*alpha*beta) = 2((m+n)^2 - (m^2 + n^2)) = 4mn. The product of the new roots is R1 * R2 = (alpha^2 - beta^2)^2 = ((alpha + beta)(alpha - beta))^2 = (alpha + beta)^2((alpha + beta)^2 - 4*alpha*beta) = (m+n)^2((m+n)^2 - 2(m^2 + n^2)) = (m+n)^2(2mn - m^2 - n^2) = (m+n)^2(-(m-n)^2) = -(m^2 - n^2)^2. Forming the new quadratic equation x^2 - (sum)x + (product) = 0 yields x^2 - 4mnx - (m^2 - n^2)^2 = 0.

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