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If b1.b2= 2(c1+c2), then at least one of the equation x2+b1X+c1=0 and X2+b2X+c2=0 will have real roots.How?

If b1.b2= 2(c1+c2), then at least one of the equation x2+b1X+c1=0 and X2+b2X+c2=0 will have real roots.How?

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question

x^2 + b_{1}x + c_{1} = 0
If this equation have real roots, then
b_{1}^{2}\geq 4c_{1}….........(1)
x^2 + b_{2}x + c_{2} = 0
If this equation have real roots, then
b_{2}^{2}\geq 4c_{2}….............(2)
And we have given
b_{1}b_{2} = 2(c_{1}+c_{2})
b_{1}^{2}b_{2}^{2} = 4(c_{1}+c_{2})^{2}
b_{1}^{2}b_{2}^{2} = 4(c_{1}^{2}+c_{2}^{2}+2c_{1}c_{2})
b_{1}^{2}b_{2}^{2} = 4(c_{1}^{2}+c_{2}^{2}-2c_{1}c_{2}+4c_{1}c_{2})
b_{1}^{2}b_{2}^{2} = 4(c_{1}-c_{2})^2+16c_{1}c_{2
b_{1}^{2}b_{2}^{2} - 16c_{1}c_{2} = 4(c_{1}-c_{2})^2
b_{1}^{2}b_{2}^{2} - 16c_{1}c_{2} = 4(c_{1}-c_{2})^2\geq 0
b_{1}^{2}b_{2}^{2} - 16c_{1}c_{2}\geq 0
b_{1}^{2}b_{2}^{2} \geq 16c_{1}c_{2}
(1)\times (2)
b_{1}^{2}b_{2}^{2} \geq 16c_{1}c_{2}
Hence proved.

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