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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!

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    $\begingroup$ 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 $(-2)^2-4(k)(3-2k)$. $\endgroup$ – player3236 Sep 28 '20 at 14:06
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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 $k$.

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

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  • $\begingroup$ 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 $\endgroup$ – Gravity098 Sep 28 '20 at 14:58
  • $\begingroup$ You are welcome, @Gravity098 $\endgroup$ – 欲しい未来 Sep 28 '20 at 15:31
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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 $k\in\{1,\frac{1}{2}\}$.

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  • $\begingroup$ Just to clarify is it positive 1/2? $\endgroup$ – Gravity098 Sep 28 '20 at 14:29
  • $\begingroup$ yes , its one by two $\endgroup$ – Lawliet Sep 28 '20 at 14:33
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$$kx^2-2x+3-2k=0$$ Since this equation has equal roots, therefore discriminant is $0$. $$\Delta=0$$ $$b^2-4ac=0$$ $$2^2-4(3-2k)(k)=0$$ $$1=(3-2k)(k)$$ $$2k^2-3k+1=0$$ $$k=\frac{1}{2},1.$$

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