Question

Find the value of $k$ for which the quadratic equation $(k-2)x^2+(2k-3)x+(5k-6)=0$ has equal roots.

Answer 1

Given the equation has equal roots

$\Rightarrow$Discriminant$=D=b^2-4ac=0$

$\Rightarrow {(2k-3)}^2-4\times (k-2)(5k-6)=0$

$\Rightarrow 4k^2+9-12k-4(5k^2-6k-10k+12)=0$

$\Rightarrow 4k^2+9-12k-4(5k^2-16k+12)=0$

$\Rightarrow 4k^2+9-12k-20k^2+64k-48=0$

$\Rightarrow -16k^2+52k-39=0$

$\Rightarrow 16k^2-52k+39=0$

Now solving the Quadratic equation by completing the square method we get:

$\Rightarrow {\left(4k\right)}^2-2\times 4k\times \frac{13}{2}+\frac{169}{4}-\frac{169}{4}+39=0$

$\Rightarrow {(4k-\frac{13}{2})}^2=\frac{169}{4}-39$

$\Rightarrow {(4k-\frac{13}{2})}^2=\frac{169-156}{4}$

$\Rightarrow {(4k-\frac{13}{2})}^2=\frac{13}{4}$

$\Rightarrow 4k-\frac{13}{2}=\pm \frac{\sqrt{13}}{2}$

$\Rightarrow 4k=\frac{13\pm \sqrt{13}}{2}$

$\therefore k=\frac{13\pm \sqrt{13}}{8}$

Thus the value of $k$ can be $\frac{13+\sqrt{13}}{8}$ or $\frac{13-\sqrt{13}}{8}$