How do you solve using the completing the square method $r^{2} + 14 r = - 13$ $r^2 + 14r = -13$ ?

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How do you solve using the completing the square method $r^{2} + 14 r = - 13$ $r^2 + 14r = -13$ ?

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Answer 1 · 2017-04-15T20:54:51

$r^{2} + 14 r = - 13$ $r^2 + 14r = -13$

$\Rightarrow r^{2} + 14 r + 13 = 0$ $implies r^2 + 14r + 13 = 0$

We could have a go at factoring that, but completing the square is the next step down the list and is what you are asking for.

It is the Quadratic Equation without having to remember a formula!

So:

$r^{2} + 14 r + 13$ $r^2 + 14r + 13$

$= r^{2} + (2 \times 7) r + 13$ $= r^2 + (color(red)(2 xx) 7)r + 13$

[Remember that: ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a + b)^2 = a^2 + color(red)(2) ab + b^2$ ]

$= {(r + 7)}^{2} - 7^{2} + 13$ $= (r + 7)^2 - 7^2 + 13$

$= {(r + 7)}^{2} - 36$ $= (r + 7)^2 - 36$

Now:

${(r + 7)}^{2} - 36 = 0$ $(r + 7)^2 - 36 color(red)(= 0)$

$\Rightarrow {(r + 7)}^{2} = 36$ $implies (r + 7)^2 = 36$

$\Rightarrow r + 7 = \sqrt{36}$ $implies r + 7 = sqrt36$

$\Rightarrow r = - 7 \pm 6 = - 1, - 13$ $implies r =- 7 pm 6 = -1, -13$

We can check that:

$(r + 1) (r + 13) = r^{2} + 13 r + r + 13 = r^{2} + 14 r + 13$ $(r+1)(r+13) = r^2 + 13r + r + 13 color(blue)(= r^2 + 14r + 13)$

Answer 2 · 2017-04-15T20:55:44

See the method outlined below:

Explanation:

First, write it in general form:

$r^{2} + 14 r + 13 = 0$ $r^2+14r+13=0$

Add bracket around the terms that contain $r$ $r$ , with the constant term outside the bracket (and factor out the $a$ $a$ coefficient if there is one):

$(r^{2} + 14 r) + 13 = 0$ $(r^2+14r)+13=0$

Now, take half of the value of $b$ $b$ (the coefficient of the term in $r$ $r$ ) and square it. Add this value and subtract it as well, within the bracket.

$(r^{2} + 14 r + 49 - 49) + 13 = 0$ $(r^2+14r+49 - 49)+13=0$

Move the subtracted ${(\frac{b}{2})}^{2}$ $(b/2)^2$ term outside the bracket. (This can require some care if there was an $a$ $a$ value (a coefficient of the $r^{2}$ $r^2$ term greater than 1.)

$(r^{2} + 14 r + 49) - 49 + 13 = 0$ $(r^2+14r+49) - 49+13=0$

By design, the part on the brackets is a perfect square:

${(r + 7)}^{2} - 36 = 0$ $(r+7)^2 -36 = 0$

It is now in standard form, so you are done.

The vertex is at (-7, -36).

Answer 3 · 2017-04-15T20:59:02

$r = {- 1, - 13}$ $r={-1,-13}$

Explanation:

$r^{2} + 14 r = - 13$ $r^2+14r=-13$

$Let us rearrange the equation.$ $"Let us rearrange the equation."$

$r^{2} + 14 r + 13 = 0$ $r^2+14r+13=0$

$let us add+36 -36$ $"let us add+36 -36"$

$r^{2} + 14 r + 36 - 36 + 13 = 0$ $r^2+14rcolor(red)(+36-36)+13=0$

$r^{2} + 14 r + 36 + 13 - 36 = 0$ $r^2+14r+36+13color(red)(-36)=0$

$r^{2} + 14 r + 49 - 36 = 0$ $color(green)(r^2+14r+49)-color(red)(36)=0$

$r^{2} + 14 r + 49 = {(r + 7)}^{2}$ $color(green)(r^2+14r+49)=(r+7)^2$

${(r + 7)}^{2} - 36 = 0$ $(r+7)^2-36=0$

${(r + 7)}^{2} - 6^{2} = 0$ $(r+7)^2-6^2=0$

$let us remember a^{2} - b^{2} = (a - b) (a + b)$ $"let us remember " a^2-b^2=(a-b)(a+b)$

$(r + 7 - 6) (r + 7 + 6) = 0$ $(r+7-6)(r+7+6)=0$

$(r + 1) (r + 13) = 0$ $(r+1)(r+13)=0$

$Either (r+1) or (r+13) is equal to zero.$ $"Either (r+1) or (r+13) is equal to zero."$

$r + 1 = 0 \Rightarrow r = - 1$ $r+1=0 rArr r=-1$

$r + 13 = 0 \Rightarrow r = - 13$ $r+13=0rArr r=-13$

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How do you solve using the completing the square method $r^{2} + 14 r = - 13$ $r^2 + 14r = -13$ ?

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How do you solve using the completing the square method r2+14r=−13r^2 + 14r = -13?

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How do you solve using the completing the square method $r^{2} + 14 r = - 13$ $r^2 + 14r = -13$ ?