# How do you find the value of the discriminant and state the type of solutions given 4k^2+5k+4=-3k?

Oct 20, 2017

$\Delta = 0$
$1$ repeated root

#### Explanation:

rearrange the equation into the form $a {k}^{2} + b k + c = 0$:

$4 {k}^{2} + 5 k + 4 = - 3 k$
$4 {k}^{2} + 5 k + 3 k + 4 = 0$
$4 {k}^{2} + 8 k + 4 = 0$

divide both sides by $4$:
${k}^{2} + 2 k + 1 = 0$

$a {k}^{2} + b k + c = 0$
${k}^{2} + 2 k + 1 = 0$

$\therefore a = 1 , b = 2 , c = 1$

$\Delta$ (discriminant) $= \left({b}^{2} - 4 a c\right)$ for a quadratic equation

here, ${b}^{2} - 4 a c = {2}^{2} - \left(4 \cdot 1 \cdot 1\right)$
$= 4 - 4$
$= 0$

$0 = 0$
since $\Delta = 0 , k$ has $1$ repeated root.

on a graph:

the parabola only meets the $x$-axis once- there is only $1$ root.

through factorisation:

${k}^{2} + 2 k + 1 = 0$
$\left(k + 1\right) \left(k + 1\right) = 0$
solving for $k$, you get
'$k = - 1$ or $k = - 1$'

the number $- 1$ is repeated, giving the name 'repeated root'.