How do you solve using the completing the square method $x^{2} - 6 x = 13$ $x^2 - 6x =13$ ?

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How do you solve using the completing the square method $x^{2} - 6 x = 13$ $x^2 - 6x =13$ ?

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Answer 1 · 2016-05-06T05:11:01

$x = 3 + \sqrt{22}$ $x=3+sqrt22$
$x = 3 - \sqrt{22}$ $x=3-sqrt22$

Explanation:

Given -

$x^{2} - 6 x = 13$ $x^2-6x=13$
Divide the coefficient of $x$ $x$ by 2 and add it to both sides

$x^{2} - 6 x + {(\frac{- 6}{2})}^{2} = 13 + + {(\frac{- 6}{2})}^{2}$ $x^2-6x+((-6)/2)^2=13 ++((-6)/2)^2$

You have a perfect square on the left hand side

$x^{2} - 6 x + 9 = 13 + 9$ $x^2-6x+9=13+9$

Rewrite it as -

${(x - 3)}^{2} = 22$ $(x-3)^2=22$

Taking square on both sides, we have

$(x - 3) = \pm \sqrt{22}$ $(x-3)=+-sqrt22$

$x = 3 + \sqrt{22}$ $x=3+sqrt22$
$x = 3 - \sqrt{22}$ $x=3-sqrt22$

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How do you solve using the completing the square method $x^{2} - 6 x = 13$ $x^2 - 6x =13$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve using the completing the square method x2−6x=13x^2 - 6x =13?

1 Answer

Explanation:

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Impact of this question

How do you solve using the completing the square method $x^{2} - 6 x = 13$ $x^2 - 6x =13$ ?