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If , then the equation (ax2 + bx – a) (dx2 + ex + f) = 0 (where a, b, c, d, e and f are real numbers) will have |
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Answer» Answer: Solution: GIVEN a,b,c in G.P. So b2 = ac Hence b = √ac ax2+2BX+c = 0 Put b = √(ac) in above equation, we get ax2+2√(ac)x+c = 0 (√(a)x+√c)2 = 0 So x = -√(c/a) Given equations ax2 + 2bx + c = 0 and dx2 + 2EX + F = 0 have a common root. So x = -√(c/a) should satisfy dx2 + 2ex + f = 0. Put x = -√(c/a) in dx2 + 2ex + f = 0. d(c/a)-2e√(c/a) + f = 0 Divide by c, we get (d/a) -2(e/√ca)+(f/c) = 0 Put b = √ac in above equation (d/a)+(f/c) = 2(e/b) So (d/a), (e/b), (f/c) are in A.P Hence option (1) is the answer. Step-by-step explanation: |
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