How do you find the discriminant and how many solutions does x^2+4x+3=4 have?

Jun 8, 2018

We get two real solutions

Explanation:

Writing your equation in the form
${x}^{2} + 4 x - 1 = 0$
the4n we get by the quadratic formula:

${x}_{1 , 2} = 2 \pm \setminus \sqrt{4 + 1}$
so we get
${x}_{1} = 2 + \sqrt{5}$

${x}_{2} = 2 - \sqrt{5}$

Jun 8, 2018

$x = - 2 \left(1 \pm \sqrt{5}\right)$

Explanation:

${x}^{2} + 4 x + 3 = 4$
$f \left(x\right) = {x}^{2} + 4 x - 1 = 0$
Discriminant D -->
$D = {d}^{2} = {b}^{2} - 4 a c = 16 + 4 = 20$ --> $d = \pm 4 \sqrt{5}$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{4}{2} \pm \frac{4 \sqrt{5}}{2} = - 2 \pm 2 \sqrt{5}$