How do you solve the equation $2 x^{2} + 10 x + 1 = 13$ $2x^2+10x+1=13$ by completing the square?

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How do you solve the equation $2 x^{2} + 10 x + 1 = 13$ $2x^2+10x+1=13$ by completing the square?

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Answer 1 · 2017-05-07T09:05:16

Put constants on one side and x terms on the other side and complete the square.

Explanation:

By subtracting 1 from both sides, we get:
$2 x^{2} + 10 x = 12$ $2x^2+10x=12$
We can simplify by dividing both sides by 2:
$x^{2} + 5 x = 6$ $x^2+5x=6$
Here we complete the square:
Since ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$ , here $a^{2}$ $a^2$ is $x^{2}$ $x^2$ and our $2 a b$ $2ab$ term is $5 x$ $5x$ , therefore our $b$ $b$ term must be $\frac{5}{2}$ $5/2$ . We complete the square by making $x^{2} + 5 x$ $x^2+5x$ into the form of the ${(a + b)}^{2}$ $(a+b)^2$ , however we also need to subtract the $b^{2}$ $b^2$ term since we had added it in to complete the square:
$(x^{2} + 5 x + {(\frac{5}{2})}^{2}) - {(\frac{5}{2})}^{2} = 6$ $(x^2+5x+(5/2)^2)-(5/2)^2=6$
${(x + \frac{5}{2})}^{2} - \frac{25}{4} = 6$ $(x+5/2)^2-25/4=6$
and simplify:
${(x + \frac{5}{2})}^{2} = \frac{49}{4}$ $(x+5/2)^2=49/4$
$x + \frac{5}{2} = \pm \sqrt{\frac{49}{4}}$ $x+5/2=+-sqrt(49/4)$
$x + \frac{5}{2} = \pm \frac{7}{2}$ $x+5/2=+-7/2$
$x = \frac{- 5 \pm 7}{2}$ $x=(-5+-7)/2$
$x = - 6 or x = 1$ $x=-6 or x=1$

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How do you solve the equation $2 x^{2} + 10 x + 1 = 13$ $2x^2+10x+1=13$ by completing the square?

1 Answer

Explanation:

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How do you solve the equation $2 x^{2} + 10 x + 1 = 13$ $2x^2+10x+1=13$ by completing the square?