Find the possible values of $k$, if the equation has equal roots.

Question

The equation $x^2+5k=kx+x+19$ has equal roots. Find the possible values of $k$.

Um having problem in rearranging the equation;

$x^2+5k-kx-x-19=0$

$x^2+k(5-x)-x-19=0$

What is the next step?

Answer 1

Try this

$$ x^2 - (k +1)x + (5k - 19) = (x - a)^2 = x^2 -2a + a^2 = 0$$

So now by inspection , $$5k - 19 = a^2 $$ $$(k+1) = 2a$$

$$ k = 2a -1 $$

$$5(2a -1) - 19 = a^2$$

$$a^2 - 10a + 24 = 0$$

$$(a -6)(a-4) = 0$$

$$a = 6 \ , \ a = 4$$

$$k = 2(6)-1 \ , \ k = 2(4)-1 $$

$$k= 11 \ , \ k = 7 $$

Answer 2

We write the quadratic equation in the form $$x^2-(k+1)x+5k-19=0$$ and it has one root if the discriminant $$\Delta=(k+1)^2-4(5k- 19)=k^2-18k+77=0$$ and the reduced discriminant of this last equation is $$\Delta'=81-77=4$$ hence the values of $k$: $$k_1=7\quad\text{and}\quad k_2=11$$