How do you solve using the completing the square method $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?

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How do you solve using the completing the square method $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?

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Answer 1 · 2017-01-20T04:55:44

Take the coefficient of the $x$ $x$ term ( $- 5$ $-5$ ) and:
1. Divide it by 2 (to get $- \frac{5}{2}$ $-5/2$ ).
2. Square this (to get $\frac{25}{4}$ $25/4$ ).
3. Add this final value to both sides.

Explanation:

Completing the square means seeking a constant term $n$ $n$ to add to $x^{2} - 5 x$ $x^2-5x$ , so that $x^{2} - 5 x + n$ $x^2-5x+n$ is a perfect square.

First, let's look at what happens when we FOIL a perfect square binomial of the form ${(x + a)}^{2}$ $(x+a)^2$ :

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

A perfect square will always have a distributed form like this.

What we notice is that if we take the coefficient of the $x$ $x$ term ( $2 a$ $2a$ ), cut it in half, and then square it, we get $a^{2}$ $a^2$ , the constant term. Thus, if given $x^{2} + 2 a x = b$ $x^2+2ax=b$ , we would complete the square by adding $a^{2}$ $a^2$ (that is, the square of half of $2 a$ $2a$ ) to both sides, giving

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

so that the trinomial on the left will be guaranteed to be a perfect square—the square of $(x + a)$ $(x+a)$ .

For this particular problem, we are given $x^{2} - 5 x = 9$ $x^2-5x=9$ . So, $- 5$ $-5$ is like our " $2 a$ $2a$ ". And if

$- 5 = 2 a$ $-5=2a$ ,

then

$a = - \frac{5}{2}$ $a=-5/2$ ,

and

$a^{2} = \frac{25}{4}$ $a^2=25/4$ .

Thus, $x^{2} - 5 x + \frac{25}{4}$ $x^2-5x+25/4$ is the completed square we seek, meaning we need to add $\frac{25}{4}$ $25/4$ to both sides:

$x^{2} - 5 x + \frac{25}{4} = 9 + \frac{25}{4}$ $x^2-5x+25/4=9+25/4$

Okay, so if this $x^{2} - 5 x + \frac{25}{4}$ $x^2-5x+25/4$ is a perfect square, what is its factored form? (Or, what is its square root?)

That's easy—remember that the factored form of $x^{2} + 2 a x + a^{2}$ $x^2+color(red)(2a)x+a^2$ is ${(x + a)}^{2}$ $(x+color(blue)a)^2$ . The $a$ $color(blue)a$ in the factor is half of the $2 a$ $color(red)(2a)$ in the trinomial. So the factored form of $x^{2} - 5 x + \frac{25}{4}$ $x^2-5x+25/4$ will be ${(x - \frac{5}{2})}^{2}$ $(x-5/2)^2$ , because $- \frac{5}{2}$ $-5/2$ is half of $- 5$ $-5$ .

We simplify both sides now to get

${(x - \frac{5}{2})}^{2} = \frac{61}{4}$ $(x-5/2)^2=61/4$

Now our LHS is a perfect square, so we can solve for $x$ $x$ by taking the square root of both sides:

$x - \frac{5}{2} = \pm \frac{\sqrt{61}}{2}$ $x-5/2=+-sqrt61 /2$

and then adding $\frac{5}{2}$ $5/2$ to both sides:

$x = \frac{5}{2} \pm \frac{\sqrt{61}}{2} = \frac{5 \pm \sqrt{61}}{2}$ $x=5/2+-sqrt61 / 2" "=" "(5+-sqrt61)/2$ .

Note:

If the coefficient on the $x^{2}$ $x^2$ term is something other than $1$ $1$ , you'll want to either factor it out or divide everything by it first, so that this method will work.

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How do you solve using the completing the square method $x^{2} - 5 x = 9$ $x^2 - 5x = 9$ ?

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Explanation:

Note:

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