How do I solve a quadratic for $k$ using the discriminant

Solution 1:

The discriminant of a quadratic $ax^2+bx+c$ is given by $\Delta=b^2-4ac.$ Moreover the quadratic has equal roots when $\Delta=0$.

In this case the quadratic is $kx^2-2x+3-2k$, so $b=-2$, $a=k$ and $c=3-2k$.

Can you end it?

Also by the quadratic equation we have $$x=\frac{2\pm\sqrt{\Delta}}{2k}=\frac{2\pm\sqrt{4-4k(3-2k)}}{2k}$$ $$=\frac{2\pm\sqrt{8k^2-12k+4}}{2k}$$ $$=\frac{2\pm2\sqrt{2k^2-3k+1}}{2k}=\frac{1\pm\sqrt{2k^2-3k+1}}{k}$$

Since you want equal roots (a repeated root), we solve $2k^2-3k+1=0$ for $k$.

Solution 2:

Once the equation has two equal roots, it means that the discriminant is equal to zero, i.e. $\Delta=0$

For $kx^2-2x+3-2k=0$

$\Delta = b^2 - 4ac \implies \Delta=4-4k(3-2k) \implies \Delta=4-12k+8k^2$

$$0=4-12k+8k^2 \implies 0=4(2k^2-3k+1) \implies 0=4(2k-1)(k-1)$$

$\therefore k=\dfrac{1}{2}; k=1 $