How do you solve by completing the square: $x^{2} = 8 x + 10$ $x^2=8x+10$ ?

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How do you solve by completing the square: $x^{2} = 8 x + 10$ $x^2=8x+10$ ?

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Answer 1 · 2018-07-07T07:09:42

$x = 4 \pm \sqrt{26}$ $x = 4+- sqrt(26)$

Explanation:

Given: $x^{2} = 8 x + 10$ $x^2 = 8x + 10$ . Solve using completing of the square.

First, group the monomials with $x$ $x$ together on the same side:

$x^{2} - 8 x = 10$ $x^2 - 8x = 10$

To complete the square, multiply the constant of the $x$ $x$ monomial by $\frac{1}{2}$ $1/2$ : $- 8 \cdot \frac{1}{2} = - \frac{8}{2} = - 4$ $" "-8 *1/2 = -8/2 = -4$

When you complete the square you end up adding a square term:
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a + b)^2 = a^2 + 2ab + b^2$

${(x - 4)}^{2} = x^{2} - 8 x + {(- 4)}^{2}$ $(x - 4)^2 = x^2 - 8x + color(red)( (-4)^2)$

This term must be also added to the right side of the equation:

${(x - 4)}^{2} = 10 + {(- 4)}^{2}$ $(x - 4)^2 = 10 + (-4)^2$

${(x - 4)}^{2} = 26$ $(x-4)^2 = 26$

Square root both sides:

$x - 4 = \pm \sqrt{26}$ $x - 4 = +- sqrt(26)$

$x = 4 \pm \sqrt{26}$ $x = 4 +- sqrt(26)$

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How do you solve by completing the square: $x^{2} = 8 x + 10$ $x^2=8x+10$ ?

1 Answer

Explanation:

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How do you solve by completing the square: $x^{2} = 8 x + 10$ $x^2=8x+10$ ?