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

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

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Answer 1 · 2018-04-07T18:32:46

$x = \pm \sqrt{26} - 5$ $x=+-sqrt26-5$

Explanation:

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

$x^{2} + 10 x = 1$ $x^2+10x=1$

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

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

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

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

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

${(x + 5)}^{2} = 26$ $(x+5)^2=26$

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

$\sqrt{{(x + 5)}^{2}} = \pm \sqrt{26}$ $sqrt((x+5)^2)=+-sqrt26$

$x + 5 = \pm \sqrt{26}$ $x+5=+-sqrt26$

Solving for $x$ $x$ yields

$x = \pm \sqrt{26} - 5$ $x=+-sqrt26-5$

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

1 Answer

Explanation:

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