How do you solve $x^{2} + 30 x - 7 = 0$ $x^2+30x-7=0$ by completing the square?

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How do you solve $x^{2} + 30 x - 7 = 0$ $x^2+30x-7=0$ by completing the square?

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Answer 1 · 2016-08-19T09:55:53

$x = 0.2315 or x = - 30.2315$ $x = 0.2315 or x = -30.2315$

Explanation:

The method using completing the square is based on:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2}$ $(x-y)^2 = x^2 - 2xy +y^2$

${(x - 6)}^{2} = x^{2} - 12 x + 36$ $(x-6)^2 = x^2 color(red)(-12)x +color(blue)(36)$ Note that: ${(\frac{- 12}{2})}^{2} = 36$ $(color(red)(-12)/2)^2 = color(blue)(36)$
This relationship always exists in squaring a binomial.

$x^{2} + 30 x - 7 = 0 7 is not the correct constant$ $x^2 +30x-7 =0" 7 is not the correct constant"$

Move the 7 to the other side and add in the correct constant on both sides.
$x^{2} + 30 x + 225 = 7 + 225 {(\frac{30}{2})}^{2} = 225$ $x^2 color(red)(+30)x color(blue)(+225)= 7 color(blue)(+225)" "(color(red)(30)/2)^2 = color(blue)225$
${(x + 15)}^{2} = 232 x^{2} + 30 x + 225 is a square$ $(x+15)^2 = 232" "x^2 +30x+225" is a square"$

$x + 15 = \pm \sqrt{232} square root both sides$ $x + 15 = +-sqrt232 " square root both sides"$

Solve for x twice:

$x = + \sqrt{232} - 15 = 0.2315$ $x = +sqrt232 -15 = 0.2315$

$x = - \sqrt{232} - 15 = - 30.2315$ $x = - sqrt232 -15 =-30.2315$

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How do you solve $x^{2} + 30 x - 7 = 0$ $x^2+30x-7=0$ by completing the square?

1 Answer

Explanation:

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Science

Math

Humanities

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How do you solve x2+30x−7=0x^2+30x-7=0 by completing the square?

1 Answer

Explanation:

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Impact of this question

How do you solve $x^{2} + 30 x - 7 = 0$ $x^2+30x-7=0$ by completing the square?