Question

If p and q are zeroes of polynomial x^2 + 2x - 1, find p - q

Answer 1

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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