Show that the equation $2(a^2+b^2)x^2+2(a+b)x+1=0$ has no real roots, when a≠b.

Given:

Given quadratic equation is $2(a^2+b^2)x^2+2(a+b)x+1=0$ and $a≠b$.

To do:

We have to show that the equation $2(a^2+b^2)x^2+2(a+b)x+1=0$ has no real roots.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2(a^2+b^2), b=2(a+b)$ and $c=1$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[2(a+b)]^2-4[2(a^2+b^2)](1)$

$D=4(a^2+2ab+b^2)-8(a^2+b^2)$

$D=4(a^2+2ab+b^2-2a^2-2b^2)$

$D=4(-a^2+2ab-b^2)$

$D=-4(a^2-2ab+b^2)$

$D=-4(a-b)^2$

$D<0$   (A negative number multiplied by a square is negative and $a≠b$)

Therefore, the given quadratic equation has no real roots.

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Updated on: 10-Oct-2022

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