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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 $2$.