How do you complete the square to solve $x^{2} - 4 x - 5 = 0$ $x^2 -4x -5=0$ ?

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How do you complete the square to solve $x^{2} - 4 x - 5 = 0$ $x^2 -4x -5=0$ ?

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Answer 1 · 2018-04-25T00:37:10

$0 = (x^2 - 4x) - 5 = (x^2 -4x + 4) -4 -5 = (x-2)^2-9 quad$ so

$(x-2)^2 = 9 or x-2 = \pm \sqrt{9} or x = 2\pm 3$ so

$x=-1 or x=5$

Answer 2 · 2018-04-25T00:45:36

$-1=x=5$

Explanation:

We can regroup the equation as:

$(x^2-4x)-5=0$

This equation is in the form:

$ax^2+bx+c=0$ where $a=1$ . (That is important.)

We divide $b$ by two.

$b=-4$

$=>b/2=-2$ square the answer.

$=>(-2)^2=4$

We add this answer inside the parenthesis

$=>(x^2-4x+4)-5=0$ Since we add four to the original equation, we need to adjust it by subtracting it by four.

$=>(x^2-4x+4)-5-4=0$

$=>(x^2-4x+4)-9=0$

Now, note that $x^2-4x+4$ is in the form

$a^2+2ab+b^2$ where $a=x$ and $b=-2$

We can rewrite this into the form $(a+b)^2$

$=>(x-2)^2-9=0$

$=>(x-2)^2=9$

$=>(x-2)=+-sqrt9$ don't forget the $+-$ !

$=>x=+-3+2$

$=>-3+2=x=3+2$

$=>-1=x=5$

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How do you complete the square to solve $x^{2} - 4 x - 5 = 0$ $x^2 -4x -5=0$ ?

2 Answers

Explanation:

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How do you complete the square to solve $x^{2} - 4 x - 5 = 0$ $x^2 -4x -5=0$ ?