Find the values of k for which the roots are real and equal in each of the following equations:

$x^2 - 2kx + 7k-12 = 0$

Given:

Given quadratic equation is $x^2 - 2kx + 7k-12 = 0$.

To do:

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

Solution:

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

$a=1, b=-2k$ and $c=7k-12$.

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

$D=(-2k)^2-4(1)(7k-12)$

$D=4k^2-4(7k-12)$

$D=4k^2-28k+48$

The given quadratic equation has real and equal roots if $D=0$.

Therefore,

$4k^2-28k+48=0$

$4(k^2-7k+12)=0$

$k^2-7k+12=0$

$k^2-4k-3k+12=0$

$k(k-4)-3(k-4)=0$

$(k-4)(k-3)=0$

$k-4=0$ or $k-3=0$

$k=4$ or $k=3$

The values of $k$ are $3$ and $4$. 

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Updated on: 10-Oct-2022

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