Algebra> quadratic...

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?

anil sinha , 12 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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.