How do you solve $x^{2} - 4 x = 10$ $x^2-4x=10$ by completing the square?

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How do you solve $x^{2} - 4 x = 10$ $x^2-4x=10$ by completing the square?

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Answer 1 · 2015-05-17T15:45:03

Solution is x=2 $\pm \sqrt{14}$ $+-sqrt14$

To make a perfect square on the left side, take half the coefficient of x and then square it. Here half of -4 is -2 and its square would be 4. Now add this number on both sides of the equation. On the left side it would now be a perfect square.

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

${(x - 2)}^{2} = 14$ $(x-2)^2 =14$

x-2= $\pm \sqrt{14}$ $+-sqrt14$

x=2 $\pm \sqrt{14}$ $+-sqrt14$

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How do you solve $x^{2} - 4 x = 10$ $x^2-4x=10$ by completing the square?

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