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If p and q are zeroes of polynomial x^2 + 2x - 1, find p - q |
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Answer» <P>Given that p and Q are zeroes of x² - 2x - 1 To find (p - q) It's known that for a quadratic equation of the form of ax² + bx + c, sum of zeroes = -b/a product of zeroes = c/a Here, in x² + 2x - 1, a = 1, b = 2 and c = -1 zeroes = p and q Hence, p + q = -(2)/1 = -2 PQ = -1/1 = -1 Now p + q = -2 → (p + q)² = (-2)² → p² + q² + 2pq = 4 SUBTRACTING 4pq from both sides → p² + q² + 2pq - 4pq = 4 - 4pq → p² + q² - 2pq = 4 - 4(-1) → (p - q)² = 4 + 4 → (p - q)² = 8 → p - q = ± √8 Or, p - q = ±2√2 |
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