# How do you find the solution to the quadratic equation x^2 - 4x -3 = 0?

Apr 14, 2018

$x = 2 \pm \sqrt{7}$

#### Explanation:

$\text{there are no whole numbers which multiply to - 3}$
$\text{and sum to - 4}$

$\text{we can solve using the method of "color(blue)"completing the square}$

$\text{the coefficient of the "x^2" term is 1}$

• " add subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - 4 x$

$\Rightarrow {x}^{2} + 2 \left(- 2\right) x \textcolor{red}{+ 4} \textcolor{red}{- 4} - 3 = 0$

$\Rightarrow {\left(x - 2\right)}^{2} - 7 = 0$

$\Rightarrow {\left(x - 2\right)}^{2} = 7$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow x - 2 = \pm \sqrt{7} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 2 \pm \sqrt{7} \leftarrow \textcolor{red}{\text{exact solutions}}$

Apr 14, 2018

x = $2 \pm \sqrt{7}$

#### Explanation:

Apply quadratic formula for this equation instead of trying to factor it out.
1/ $\left(\frac{- b \pm \sqrt{{\left(b\right)}^{2} - 4 \left(a\right) \left(c\right)}}{2 \left(a\right)}\right)$

2/ $\left(\frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 3\right)}}{2 \left(1\right)}\right)$

3/ $\left(\frac{4 \pm \sqrt{16 + 12}}{2}\right)$

4/ $\left(\frac{4 \pm 2 \sqrt{7}}{2}\right)$ ( 2 cancel out)

5/ x = $2 \pm \sqrt{7}$

Apr 14, 2018

$x = 2 + \sqrt{7} \mathmr{and} x = 2 - \sqrt{7}$

#### Explanation:

Here,

${x}^{2} - 4 x - 3 = 0$

$\implies {x}^{2} - 4 x + 4 - 7 = 0$

$\implies {\left(x - 2\right)}^{2} = 7 = {\left(\sqrt{7}\right)}^{2}$

$\implies x - 2 = \pm \sqrt{7}$

$\implies x = 2 \pm \sqrt{7}$

OR

$a {x}^{2} + b x + c = 0 \implies a = 1 , b = - 4 , c = - 3$

$\triangle = {b}^{2} - 4 a c = {\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 3\right)$

$\implies \triangle = 16 + 12 = 28 = 4 \times 7$

$\sqrt{\triangle} = 2 \sqrt{7}$

So,

$x = \frac{- b \pm \sqrt{\triangle}}{2 a}$

$x = \frac{4 \pm 2 \sqrt{7}}{2 \left(1\right)}$

$x = 2 \pm \sqrt{7}$