How do you solve using the completing the square method $x^{2} + 10 x - 4 = 0$ $x^2+10x-4=0$ ?

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How do you solve using the completing the square method $x^{2} + 10 x - 4 = 0$ $x^2+10x-4=0$ ?

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Answer 1 · 2016-08-01T10:37:14

$x = \pm \sqrt{29} - 5$ $x=+-sqrt29-5$

Explanation:

You can express your quadratic equation as:
$x^{2} + 10 x - 4 = {(x + a)}^{2} + b$ $x^2+10x-4=(x+a)^2+b$

${(x + a)}^{2} + b = x^{2} + 2 a x + a^{2} + b$ $(x+a)^2+b=x^2+2ax+a^2+b$

You will find $2 a x = 10 x$ $2ax=10x$ hence $a = 5$ $a=5$
$a^{2} + b = - 4$ $a^2+b=-4$
$5^{2} + b = - 4$ $5^2+b=-4$
$b = - 29$ $b=-29$

Now you have: ${(x + 5)}^{2} - 29 = 0$ $(x+5)^2-29=0$
${(x + 5)}^{2} = 29$ $(x+5)^2=29$
$(x + 5) = \pm \sqrt{29}$ $(x+5)=+-sqrt29$
$thereforex=+-sqrt29-5$

Answer 2 · 2016-08-01T10:43:15

$x= -5+-sqrt29$

Explanation:

$x^2+10x-4=0 or (x+5)^2 -25-4=0 or (x+5)^2=29 or (x+5)=+-sqrt29 or x= -5+-sqrt29$ [Ans]

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How do you solve using the completing the square method $x^{2} + 10 x - 4 = 0$ $x^2+10x-4=0$ ?

2 Answers

Explanation:

Explanation:

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How do you solve using the completing the square method x^2+10x-4=0x2+10x−4=0x^2+10x-4=0?

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How do you solve using the completing the square method $x^{2} + 10 x - 4 = 0$ $x^2+10x-4=0$ ?