How do you solve using the completing the square method $4 v^{2} + 16 v = 65$ $4v^2+ 16v=65$ ?

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Answer 1 · 2016-05-20T14:48:55

Follow the steps below to get to $v = - 2 + \frac{9}{2} = \frac{5}{2}$ $v=-2+9/2=5/2$ ,
$v = - 2 - \frac{9}{2} = - \frac{13}{2}$ $v=-2-9/2=-13/2$

To complete the square, we first want the set up we have in this problem, that is v terms on one side and the constant on the other.

So first we want a clean $v^{2}$ $v^2$ term, so we'll divide through by its coefficient:

$4 v^{2} + 16 v = 65$ $4v^2+16v=65$

$v^{2} + 4 v = \frac{65}{4}$ $v^2+4v=65/4$

Now we take the $v$ $v$ coefficient, divide by 2, then square it and add it to both sides:

${(\frac{4}{2})}^{2} = 2^{2} = 4$ $(4/2)^2=2^2=4$

$v^{2} + 4 v + 4 = \frac{65}{4} + 4$ $v^2+4v+4=65/4+4$

Now we convert the left side of the equation to a square (and simplify the right):

${(v + 2)}^{2} = \frac{65}{4} + \frac{16}{4} = \frac{81}{4}$ $(v+2)^2=65/4+16/4=81/4$

Now take the square root of both sides:

$v + 2 = \pm \sqrt{\frac{81}{4}} = \pm \frac{9}{2}$ $v+2=+-sqrt(81/4)=+-9/2$

And finally solve for $v$ $v$ :

$v = - 2 \pm \frac{9}{2}$ $v=-2+-9/2$

$v = - 2 + \frac{9}{2} = \frac{5}{2}$ $v=-2+9/2=5/2$
$v = - 2 - \frac{9}{2} = - \frac{13}{2}$ $v=-2-9/2=-13/2$

How do you solve using the completing the square method 4v^2+ 16v=654v2+16v=654v^2+ 16v=65?