How do you solve the quadratic equation by completing the square: $x^{2} - 10 x = 1$ $x^2 - 10x = 1$ ?

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How do you solve the quadratic equation by completing the square: $x^{2} - 10 x = 1$ $x^2 - 10x = 1$ ?

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Answer 1 · 2015-08-05T23:14:31

$x_{1, 2} = 5 \pm \sqrt{26}$ $x_(1,2) = 5 +- sqrt(26)$

Explanation:

To solve this quadratic by completing the square you need to use the coefficient of the $x$ $x$ -term to help you determine what terms needs to be added to both sides of the equation so that the left side of the equation can be written as the square of a binomial.

More specifically, divide said coefficient by $2$ $2$ , square the result, and add it to both sides of the equation

${(- \frac{10}{5})}^{2} = 25$ $(-10/5)^2 = 25$

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

The left side of the equation can now be written as

$x^{2} - 2 \cdot (5) \cdot x + {(5)}^{2} = {(x - 5)}^{2}$ $x^2 - 2 * (5) * x + (5)^2 = (x-5)^2$

This means that you have

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

Take the square root from both sides of the equation to get

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

$x - 5 = \pm \sqrt{26} \Rightarrow x_{1, 2} = 5 \pm \sqrt{26}$ $x-5 = +- sqrt(26) implies x_(1,2) = 5 +- sqrt(26)$

The two solutions to the quadratic will be

$x_{1} = 5 + \sqrt{26}$ $x_1 = color(green)(5 + sqrt(26))$ and $x_{2} = 5 - \sqrt{26}$ $x_2 = color(green)(5 - sqrt(26))$

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How do you solve the quadratic equation by completing the square: $x^{2} - 10 x = 1$ $x^2 - 10x = 1$ ?

1 Answer

Explanation:

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