Using the Discriminant to find the value of 'k'.

Question

Find the value(s) of $k$ for which the equation, $(x+2)(x+k)=-1$, has equal roots.

(I cannot get the two values as stated in the answer $k=0$ and $k=4$. My final line of working doesn't seem to factorize, it is $k^2-4k+8=0$)

Answer 1

We have: $(x+2)(x+k)=-1$

$\Rightarrow x^{2}+kx+2x+2k=-1$

$\Rightarrow x^{2}+(2+k)\hspace{1 mm}x+2k+1=0$

We know that the equation has equal, or repeated, roots:

$\Rightarrow b^{2}-4ac=0$

$\Rightarrow (2+k)^{2}-4(1)(2k+1)=0$

$\Rightarrow 4+4k+k^{2}-8k-4=0$

$\Rightarrow k^{2}-4k=0$

$\Rightarrow k\hspace{1 mm}(k-4)=0$

Using the null factor law:

$\Rightarrow k=0,4$

Answer 2

the discrimnant is given by $$k^2-4k$$ and we get $$k(k-4)=0$$ if $$k=0$$ or $$k=4$$

Answer 3

$$(x+2)(x+k)=-1\implies x^2+(k+2)x+2k+1=0$$

The discriminant of the above must be zero , thus:

$$(k+2)^2-4(2k+1)=k^2-4k=0\iff k(k-4)=0\;\ldots$$