Question

If the roots of the equation $x^2-2(5+2k)x+3(7+10k)=0$ are real and equal, then find the value of $k$.

Answer 1

The given equation $x^2-2(5+2k)x+3(7+10k)=0$ is in the form of $a{x}^2+bx+c=0$
Where $a=1,b=-2(5+2k),c=3(7+10k)$
For the equation to have real and equal roots, the condition is $D=b^2-4ac=0$
$\Rightarrow (-2(5+2k){)}^2-4(1\left)\right(3(7+10k))=0$
$\Rightarrow 4(5+2k{)}^2-12(7+10k)=0$
$\Rightarrow 25+4k^2+20k-21-30k=0$
$\Rightarrow 4k^2-10k+4=0$
$\Rightarrow 2k^2-5k+2=0$ [dividing by 2]
Now, solving for $k$ by factorization, we have
$\Rightarrow 2k^2-4k-k+2=0$
$\Rightarrow 2k(k-2)-1(k-2)=0$
$\Rightarrow (k-2)(2k-1)=0,$
$k=2$ and $k=\frac{1}{2}$,
So, the value of $k$ can either be $2$ or $\frac{1}{2}$