Question

Let \( p \) and \( q \) be two positive numbers such that \( p + q = 2 \) and \( p^4 + q^4 = 272 \). Then \( p \) and \( q \) are roots of the equation :

• A. $x^2 - 2x + 136 = 0$
• B. $x^2 - 2x + 16 = 0$
• C. $x^2 - 2x + 8 = 0$
• D. $x^2 - 2x + 2 = 0$

Hint:

Use the identity $a^4 + b^4 = ((a+b)^2 - 2ab)^2 - 2(ab)^2$ to quickly relate the sum and product of roots to higher powers.

Answer 1

The Correct Option is B

Step 1: We have $p+q = 2$. Let $pq = P$.
Step 2: $p^2 + q^2 = (p+q)^2 - 2pq = 4 - 2P$.
Step 3: $p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2 = (4 - 2P)^2 - 2P^2 = 272$.
Step 4: $16 + 4P^2 - 16P - 2P^2 = 272 \Rightarrow 2P^2 - 16P - 256 = 0$.
Step 5: $P^2 - 8P - 128 = 0 \Rightarrow (P-16)(P+8) = 0$.
Step 6: Since $p, q>0$, $pq = P$ must be positive. Thus $P = 16$.
Step 7: Equation is $x^2 - (p+q)x + pq = 0 \Rightarrow x^2 - 2x + 16 = 0$.