Find the values of k for which the following equations have real roots
$kx(x-2\sqrt5) + 10 = 0$

Given:

Given quadratic equation is $kx(x-2\sqrt5)+10=0$.

To do:

We have to find the values of k for which the roots are real.

Solution:

$kx(x-2\sqrt5)+10=0$

$kx^2-(2\sqrt5)kx+10=0$

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

$a=k, b=-(2\sqrt5)k$ and $c=10$.

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

$D=(-2\sqrt{5}k)^2-4(k)(10)$

$D=4(5)k^2-40k$

$D=20k^2-40k$

The given quadratic equation has real roots if $D≥0$.

Therefore,

$20k^2-40k≥0$

$20k(k-2)≥0$

$20k≥0$ and $k-2≥0$

$k≥0$ and $k≥2$

This implies,

$k≥2$

The value of k is greater than or equal to $2$.

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Updated on: 10-Oct-2022

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