Question

X2+2(3a+5)x+2(9a2+25)=0 show the roots r complx​

Answer 1

We assume that a is real.

The discriminant of the quadratic equation x^2 +2(3a+5)x +2(9a^2 +25) = 0 is

∆ = [2(3a+5)]^2 - 4×1×2(9a^2 +25] =

[4(9a^2 +30a+25] - [72a^2 +200] =

-36a^2 +120a -100

= -4[9a^2 -30a+25] = -(3a-5)^2 < 0, unless a=(5/3), when it is 0. Hence the roots are a conjugate pair of complex numbers if a is not equal to 5/3, and for a=(5/3), there is a repeated root -(3a+5) = -10.

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