Using the discriminant to find the value of k.

Question

Question:-

$x^2-4x-1=2k(x-5)$ has two equal roots. Calculate the possible values of $k$.

I know that that must mean the discriminant must equal $0$. So I found:

$b = (-2k-4)$

$a = 1$

$c = (10k-1).$

Yet when I input this into $b^2 - 4ac$ I always get an incorrect answer. The correct answers are supposed to be $k = 1$ or $k = 5$.

Can someone explain where I went wrong? Thanks.

Welcome to Mathematics Stack Exchange. What did you get for $b^2-4ac$ in terms of $k$? — J. W. Tanner, Sep 17 '21 at 18:59
I always got really strange irrational numbers. The error was that I failed to factorise out -1 from my value for b. — tacticalcamel, Sep 17 '21 at 19:41

score 3 · Accepted Answer · answered Sep 17 '21 at 18:58

3

You have$$x^2-4x-1=2k(x-5)\iff x^2-(4+2k)x+10k-1=0.$$And\begin{align}\bigl(-(4+2k)\bigr)^2-4(10k-1)=0&\iff4(k^2-6k+5)=0\\&\iff k=1\vee k=5.\end{align}

answered Sep 17 '21 at 18:58

José Carlos Santos

427,504

Thanks, the error I made was not factoring out -1 from (-4-2k) – tacticalcamel Sep 17 '21 at 19:38

score 0 · Answer 2 · answered Sep 19 '21 at 03:34

You could also approach the problem this way. If the equation has two equal roots, then the quadratic polynomial $ \ x^2 \ - \ 2·(k+2) x \ + \ (10k - 1) \ $ is a binomial-square. This requires that $$ (k+2)^2 \ = \ 10k - 1 \ \ \Rightarrow \ \ k^2 \ + \ 4k \ + \ 4 \ - \ 10k \ + \ 1 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ k^2 \ - \ 6k \ + \ 5 \ \ = \ \ (k \ - \ 1) · (k \ - \ 5 ) \ \ = \ \ 0 \ \ . $$

Using the discriminant to find the value of k.

2 Answers2