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

Question

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

3 Answers

Impact of this question

2628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-19T14:50:57

$r_{1} = \frac{5 + 2 \sqrt{3}}{2}$ $r_1=(5+2sqrt(3))/2$ and $r_{2} = \frac{5 - 2 \sqrt{3}}{2}$ $r_2=(5-2sqrt(3))/2$

Explanation:

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

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

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

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

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

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

Hence $r_{1} = \frac{5 + 2 \sqrt{3}}{2}$ $r_1=(5+2sqrt(3))/2$ and $r_{2} = \frac{5 - 2 \sqrt{3}}{2}$ $r_2=(5-2sqrt(3))/2$

Answer 2 · 2017-10-26T18:44:46

$r = 4.232 or r = 0.768$ $r = 4.232" "or r = 0.768$

Explanation:

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

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

$4 r^{2} - 20 r + 13 = 0 \leftarrow$ $4r^2 -20r +13=0" "larr$ now $\div 4$ $div4$

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

$r^{2} - 5 r + ? ? ? = - \frac{13}{4} + ? ? ? \leftarrow$ $r^2 -5r+color(blue)(???) = -13/4+color(blue)(???)" "larr$ add ${(\frac{b}{2})}^{2}$ $color(blue)((b/2)^2)$ to both sides

$r^{2} - 5 r + {(- \frac{5}{2})}^{2} = - \frac{13}{4} + {(- \frac{5}{2})}^{2}$ $r^2 -5r +color(blue)((-5/2))^2 = -13/4 +color(blue)((-5/2)^2)$

${(r - \frac{5}{2})}^{2} = - \frac{13}{4} + \frac{25}{4} = \frac{12}{4} = 3$ $" "(r-5/2)^2 = -13/4+color(blue)(25/4) = 12/4 =3$

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

$r - \frac{5}{2} = \pm \sqrt{3}$ $r-5/2 = +-sqrt3$

$r = + \sqrt{3} + 2.5 or r = - \sqrt{3} + 2.5$ $r = +sqrt3+2.5" "or r = -sqrt3+2.5$

$r = 4.232 or r = 0.768$ $r = 4.232" "or r = 0.768$

Answer 3 · 2017-10-26T19:12:50

$r = \frac{5}{2} - \sqrt{3} or r = \frac{5}{2} + \sqrt{3}$ $r= 5/2-sqrt3 or r=5/2+sqrt3$

Explanation:

$- 4 r^{2} + 21 r = r + 13$ $-4r^2 +21r =r+13$
Let's start by subtracting $r + 13$ $color(red)(r+13)$ from both sides
$-4r^2 +21r - color(red)(r+13)=cancel(r+13)cancelcolor(red)(-r-13)$
$4r^2 + 21r -r - 13= 0$
$4r^2 +20r -13=0$
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.
$r = ((-b)+-sqrt(b^2-4ac))/(2a)$
Replace the numbers
$r = ((-20)+-sqrt((20)^2-4(-4)(-13)))/((2)(-4))$
$r=((-20)+-sqrt(192))/-8$
$r= 5/2 -sqrt3 or r=5/2 +sqrt3$

Science

Math

Humanities

... and beyond

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

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve -4r^2+21r=r+13−4r2+21r=r+13-4r^2+21r=r+13 by completing the square?

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

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