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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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