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

Picture of the question

Given that the equation $$kx^2-2x+3-2k=0$$ has equal roots, find the possible values of the constant $$k$$.

So far I have only attempted to factor $$-2$$ which I'm not sure is even right. I understand I need to use the discriminant later on in the Question. Completely lost, any guidance would be helpful. If you post the solution please show it step by step.

Thanks!

• You cannot factor $-2$ from the equation, but you can use the discriminant right here. Write the equation as $kx^2-2x+(3-2k)$, and we see that the discriminant is $(-2)^2-4(k)(3-2k)$. – player3236 Sep 28 '20 at 14:06

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 $$k$$.

• Yep. Thank you very much for your clear guidance. Much appreciated. – Gravity098 Sep 28 '20 at 14:52

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

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

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

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

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

• Thank The main issue I was facing was finding the "c" in ax^2 +bx+c. Your way of going about solving the quadratic using normal expanding was also of much use . Thank you very much – Gravity098 Sep 28 '20 at 14:58
• You are welcome, @Gravity098 – 欲しい未来 Sep 28 '20 at 15:31

It is said equal roots, so $$\sqrt{b^2-4ac} = 0$$ \begin{align*} \sqrt{4-4\cdot k\cdot (3-2k)} & = 0\\ 2\sqrt{1-3k + 2k^2} &=0 \\ \end{align*}

Now, solve for $$2k^2-3k+1 = 0$$ which when solved using $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$, you will get $$k\in\{1,\frac{1}{2}\}$$.

• Just to clarify is it positive 1/2? – Gravity098 Sep 28 '20 at 14:29
• yes , its one by two – Lawliet Sep 28 '20 at 14:33

$$kx^2-2x+3-2k=0$$ Since this equation has equal roots, therefore discriminant is $$0$$. $$\Delta=0$$ $$b^2-4ac=0$$ $$2^2-4(3-2k)(k)=0$$ $$1=(3-2k)(k)$$ $$2k^2-3k+1=0$$ $$k=\frac{1}{2},1.$$