How do you solve -4r^2+21r=r+13 by completing the square?

Oct 19, 2017

${r}_{1} = \frac{5 + 2 \sqrt{3}}{2}$ and ${r}_{2} = \frac{5 - 2 \sqrt{3}}{2}$

Explanation:

$- 4 {r}^{2} + 21 r = r + 13$

$4 {r}^{2} - 21 r + r + 13 = 0$

$4 {r}^{2} - 20 r + 13 = 0$

$4 {r}^{2} - 20 r + 25 - 12 = 0$

${\left(2 r - 5\right)}^{2} - {\left(2 \sqrt{3}\right)}^{2} = 0$

$\left(2 r - 5 + 2 \sqrt{3}\right) \cdot \left(2 r - 5 - 2 \sqrt{3}\right) = 0$

Hence ${r}_{1} = \frac{5 + 2 \sqrt{3}}{2}$ and ${r}_{2} = \frac{5 - 2 \sqrt{3}}{2}$

Oct 26, 2017

$r = 4.232 \text{ } \mathmr{and} r = 0.768$

Explanation:

$- 4 {r}^{2} + 21 r = r + 13 \text{ } \leftarrow$ make ${r}^{2}$ term positive:

$0 = 4 {r}^{2} - 21 r + r + 13 \text{ } \leftarrow$simplify

$4 {r}^{2} - 20 r + 13 = 0 \text{ } \leftarrow$ now $\div 4$

${r}^{2} - 5 r + \frac{13}{4} = 0 \text{ } \leftarrow$ move the constant to the RHS

r^2 -5r+color(blue)(???) = -13/4+color(blue)(???)" "larr add $\textcolor{b l u e}{{\left(\frac{b}{2}\right)}^{2}}$ to both sides

${r}^{2} - 5 r + {\textcolor{b l u e}{\left(- \frac{5}{2}\right)}}^{2} = - \frac{13}{4} + \textcolor{b l u e}{{\left(- \frac{5}{2}\right)}^{2}}$

$\text{ } {\left(r - \frac{5}{2}\right)}^{2} = - \frac{13}{4} + \textcolor{b l u e}{\frac{25}{4}} = \frac{12}{4} = 3$

$\text{ "(r-5/2)^2 =3" } \leftarrow$ find square root of both sides

$r - \frac{5}{2} = \pm \sqrt{3}$

$r = + \sqrt{3} + 2.5 \text{ } \mathmr{and} r = - \sqrt{3} + 2.5$

$r = 4.232 \text{ } \mathmr{and} r = 0.768$

Oct 26, 2017

$r = \frac{5}{2} - \sqrt{3} \mathmr{and} r = \frac{5}{2} + \sqrt{3}$

Explanation:

$- 4 {r}^{2} + 21 r = r + 13$
Let's start by subtracting $\textcolor{red}{r + 13}$ from both sides
$- 4 {r}^{2} + 21 r - \textcolor{red}{r + 13} = \cancel{r + 13} \cancel{\textcolor{red}{- r - 13}}$
$4 {r}^{2} + 21 r - r - 13 = 0$
$4 {r}^{2} + 20 r - 13 = 0$
Now we can use the quadratic formula
We have:
$a = - 4$
$b = 20$
$c = - 13$
Note: The quadratic formula is a formula that is very important and I recommend you to memorize it. At least understand how and when to use it.
$r = \frac{\left(- b\right) \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Replace the numbers
$r = \frac{\left(- 20\right) \pm \sqrt{{\left(20\right)}^{2} - 4 \left(- 4\right) \left(- 13\right)}}{\left(2\right) \left(- 4\right)}$
$r = \frac{\left(- 20\right) \pm \sqrt{192}}{-} 8$
$r = \frac{5}{2} - \sqrt{3} \mathmr{and} r = \frac{5}{2} + \sqrt{3}$