Question

Solve x2 + 10x + 7 = 0 by completing the square.


Which equation is used in the process?


(x+ 10)2= 93
(x+ 5)2= 32
(x+ 5)2= 18

Answer 1

To complete the square, you start by collecting all the powers of the variable on the left side, and all the loose numbers on the right side.

\[x^2+10x + 7 = 0\]\[x^2+10x= -7\]
If there's a coefficient other than 1 on the squared term, you would divide through by that now. There isn't.

Take half the value of the coefficient of the variable, square it, and add to both sides. This will allow you to rewrite the left hand side as (variable + half the coefficient)^2.

That should be enough to find the answer here.

Answer 2

an almost sneaky way but probably longer way is to expand the binomial
then subtrract the right hand side value lol. BUt you'd have to do it for all choices.

And it is a good way to check if you got it right.

Answer 3

I cant get it for some reason

Answer 4

https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/completing-the-square
explains it perfectly.

Answer 5

keep in mind that "perfect square trinomials" have the form of
\(\huge a^2\pm 2ab+b^2\)

Answer 6

so thus far you have \(x^2 + 10x + 7 = 0 \implies x^2+2(x)(5)+7 = 0\)
notice the middle term, 2x(5), so what you need for the 3rd term, is really THAT squared, or \(5^2\)
now if you ADD that, you have to also substract it too, because just to remind you
what we're really doing is borrowing from "zero" 0, so if you add an amount, you'd also need to substract it

Answer 7

\[x^2+10x = -7\]Half of 10 is 5, so we add \(5^2\) to both sides:
\[x^2+10x +5^2 = -7+5^2\]\[x^2+10x+25 = -7 + 25\]\[x^2+10x+25=18\]
so that's enough to pick an answer, I think :-)

Answer 8

$$\large {
x^2 + 10x + 7 = 0\\
x^2+2(x)(5)+5^2-5^2+7=0\\
x^2+10x+5^2-5^2+7=0\\
(x-5)^2-25+7=0
}
$$

Answer 9

woops + rather, darn

Answer 10

but to continue, the left hand side turns into \[(x+10/2)^2 = 18\]\[(x+5)^2=18\]Take the square root of both sides
\[(x+5) = \pm3\sqrt{2}\] We need to take both positive and negative square root to get both solutions
Now solve for \(x\)
\[x+5=3\sqrt{2}\]\[x = -5+3\sqrt{2}\]
\[x+5=-3\sqrt{2}\]\[x = -5-3\sqrt{2}\]

so solutions are \[x = -5\pm3\sqrt{2}\]

Answer 11

x^2 + 10x + 7 = 0\\
x^2+2(x)(5)+5^2-5^2+7=0\\
x^2+10x+5^2-5^2+7=0\\
(x+5)^2-25+7=0

Answer 12

shoot

Answer 13

$$ \large {
x^2 + 10x + 7 = 0\\
x^2+2(x)(5)+5^2-5^2+7=0\\
x^2+10x+5^2-5^2+7=0\\
(x+5)^2-25+7=0
}
$$