Question
If p = 34 - q - r and pq + r(q +
p) = 289, then find the value of (p² + q² + r²).
Solution
We have, p = 34 - q - r So, p + q + r = 34 And, pq + r(q + p) = 289 Or, pq + qr + pr = 289 Using, (p + q + r)² = p² + q² + r² + 2(pq + qr + pr) So, 34² = (p² + q² + r²) + 2 × 289 Or, 1156 = (p² + q² + r²) + 578 Or, p² + q² + r² = 1156 - 578 Or, p² + q² + r² = 578
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