Question

If the roots of the equations ax² + 2bx + c = 0 and bx²-2√acx+b=0 are simultaneously real, then prove that b²= ac i.e. a,b,c are in G.P.

Answer 1

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Answer 2

Answer:

b² = ac

Explanation:

(i) ax² + 2bx + c = 0

On comparing with ax² + bx + c = 0, we get

a = a, b = 2b, c = c

Given that the equation has real roots.

b² - 4ac ≥ 0

⇒ (2b)² - 4ac ≥ 0

⇒ 4b² - 4ac ≥ 0

⇒ b² - ac ≥ 0

⇒ b² ≥ ac

(ii) bx² - 2√acx + b² = 0

On comparing with ax² + bx + c = 0, we get

a = b, b = -2√ac, c = b

Given that the equation has real roots.

b² - 4ac ≥ 0

⇒ (-2√ac)² - 4(b)(b) ≥ 0

⇒ 4ac - 4b² ≥ 0

⇒ ac - b² ≥ 0

⇒ ac ≥ b²

From (i) & (ii), we get

b² = ac.

Hope it helps!