How do you solve using the completing the square method $x^{2} - x = 30$ $x^2 − x = 30$ ?

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How do you solve using the completing the square method $x^{2} - x = 30$ $x^2 − x = 30$ ?

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Answer 1 · 2016-08-06T05:40:16

$x = - 5, 6$ $x=-5,6$

Explanation:

$x^{2} - x = 30$ $x^2-x=30$

1) Check the constant term is on the right side if not bring it to the right side .
2) Check the coefficient of x^2 is 1 if not Make the coefficient of x^2 as 1

$x^{2} - x = 30$ $x^2-x=30$

Add both side ${(c o e f f i c i e n t o f \frac{x}{2})}^{2}$ $(coefficient of x/2)^2$

Coefficient of x is -1 so add ${(- \frac{1}{2})}^{2}$ $(-1/2)^2$ , both side

$x^{2} - x + {(\frac{1}{2})}^{2} = 30 + {(\frac{1}{2})}^{2}$ $x^2-x+(1/2)^2=30 +(1/2)^2$ use the identity ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$

$x^{2} - x + {(\frac{1}{2})}^{2} = {(x - \frac{1}{2})}^{2}$ $x^2-x+(1/2)^2=(x-1/2)^2$

${(x - \frac{1}{2})}^{2} = 30 + \frac{1}{4}$ $(x-1/2)^2=30+1/4$

${(x - \frac{1}{2})}^{2} = \frac{121}{4}$ $(x-1/2)^2=121/4$
squaring on both side

$(x - \frac{1}{2}) = \pm \sqrt{\frac{121}{4}}$ $(x-1/2)=+-sqrt(121/4)$

$(x - \frac{1}{2}) = \pm \frac{11}{2}$ $(x-1/2)=+-11/2$

$x = \frac{1}{2} + \frac{11}{2}, x = \frac{1}{2} - \frac{11}{2}$ $x=1/2+11/2, x=1/2-11/2$
$x = \frac{12}{2} or x = - \frac{10}{2}$ $x=12/2 or x=-10/2$

$x = - 5, 6$ $x=-5,6$

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How do you solve using the completing the square method $x^{2} - x = 30$ $x^2 − x = 30$ ?

1 Answer

Explanation:

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Science

Math

Humanities

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How do you solve using the completing the square method x2−x=30x^2 − x = 30?

1 Answer

Explanation:

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How do you solve using the completing the square method $x^{2} - x = 30$ $x^2 − x = 30$ ?