How do you solve $x^2+10x-1=0$ by completing the square?

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Answer 1 · 2018-04-07T18:32:46

$x=+-sqrt26-5$

Isolate all terms involving $x$ on one side and move the constant to the other side.

$x^2+10x=1$

Now, the original quadratic, $x^2+10x-1,$ is in the form $ax^2+bx+c$ where $a=1, b=10, c=-1$ .

To complete the square, we first want to add $(b/2)^2$ to each side.

$(b/2)^2=(10/2)^2=5^2=25$ , so we add $25$ to each side.

$x^2+10x+25=1+25$

The left side needs to be factored. Fortunately, since we added $(b/2)^2$ to each side, the factored form will simply be $(x+b/2)^2=(x+5)^2$

$(x+5)^2=26$

Now, take the root of both sides, accounting for positive and negative answers on the right.

$sqrt((x+5)^2)=+-sqrt26$

$x+5=+-sqrt26$

Solving for $x$ yields

$x=+-sqrt26-5$

How do you solve x^2+10x-1=0x^2+10x-1=0 by completing the square?