How do you solve $3 x^{2} - 2 x - 3 = 0$ $3x^2-2x-3=0$ by completing the square?

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How do you solve $3 x^{2} - 2 x - 3 = 0$ $3x^2-2x-3=0$ by completing the square?

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Answer 1 · 2017-07-24T04:47:15

$x = \pm \sqrt{\frac{10}{9}} + \frac{1}{3}$ $x = +-sqrt(10/9) + 1/3$ , or $\approx 1.39$ $~~ 1.39$ and $\approx - 0.72$ $~~-0.72$

Explanation:

First things first, in order to complete the square, the leading coefficient ( $3 x^{2} - 2 x - 3$ $color(red)(3)x^2-2x-3$ ) must be $1$ $1$ . To do that, we need to factor out a $3$ $3$ from the equation.

$3 x^{2} - 2 x - 3 = 0$ $3x^2-2x-3 =0$

$3 (x^{2} - \frac{2}{3} x - 1) = 0$ $3(x^2-2/3x-1)=0$

Now we have the beginning. Completing the square can look scary, but it's really just a process, and if you understand the steps, it becomes pretty simple.

The first step is getting the leading coefficient to $1$ $1$ . After that, we need to take the middle term, $- \frac{2}{3}$ $-2/3$ and "do some stuff with it" (you'll see in a minute)

So, we'll take $(- \frac{2}{3})$ $(-2/3)$ and divide it by $2$ $2$ , which gives us $- \frac{1}{3}$ $-1/3$ . Now we square the solution, which equals $\frac{1}{9}$ $1/9$ .

We did all of this because we needed to find the value that will make the left side of our equation, $x^{2} - \frac{2}{3} x - 1 = 0$ $x^2-2/3x-1 =0$ , a perfect square, which is $\frac{1}{9}$ $1/9$ .

Now that we have our missing value, we need to add it to our equation.

$x^{2} - \frac{2}{3} x - 1 + \frac{1}{9} = 0$ $x^2-2/3x-1 + color(red)(1/9) " "=0$

But wait!! We can't just add a random number into an equation! An equation is all about balance (the root word is equal). You can't just introduce a new value. But.... if you add $500$ $500$ , and then immediately subtract $500$ $500$ , the final result is $0$ $0$ .

So, if we add $\frac{1}{9}$ $color(red)(1/9)$ , and then subtract $\frac{1}{9}$ $color(red)(1/9)$ , then technically we haven't changed anything

$x^{2} - \frac{2}{3} x - 1 + \frac{1}{9} + - \frac{1}{9} = 0$ $x^2-2/3x-1 + color(red)(1/9) + color(red)(-1/9)=0$

Let's re-order this:

$x^{2} - \frac{2}{3} x + \frac{1}{9} - 1 - \frac{1}{9} = 0$ $color(green)(x^2-2/3x+1/9) color(blue)(-1 -1/9)=0$

$x^{2} - \frac{2}{3} x + \frac{1}{9}$ $color(green)(x^2-2/3x+1/9)$ is a perfect square (that was the whole point of all of this, after all). Let's factorise it.

$x^{2} - \frac{2}{3} x + \frac{1}{9} = {(x - \frac{1}{3})}^{2}$ $color(green)(x^2-2/3x+1/9) =color(green)((x-1/3)^2)$

Let's simplify this: $- 1 - \frac{1}{9}$ $color(blue)(-1 -1/9)$ equals $- \frac{10}{9}$ $color(blue)(-10/9)$

So, now we have ${(x - \frac{1}{3})}^{2} - \frac{10}{9} = 0$ $color(green)((x-1/3)^2)color(blue)(-10/9)=0$

$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$ $* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *$

Now, let's solve this puppy!

${(x - \frac{1}{3})}^{2} - \frac{10}{9} = 0$ $(x-1/3)^2-10/9=0$

add $\frac{10}{9}$ $10/9$ to both sides

${(x - \frac{1}{3})}^{2} = \frac{10}{9}$ $(x-1/3)^2 = 10/9$

take a square root of both sides

$\sqrt{{(x - \frac{1}{3})}^{2}} = \pm \sqrt{\frac{10}{9}}$ $sqrt((x-1/3)^2) = +-sqrt(10/9)$

$x - \frac{1}{3} = \pm \sqrt{\frac{10}{9}}$ $x-1/3 = +-sqrt(10/9)$

add $\frac{1}{3}$ $1/3$ to both sides

$x = \pm \sqrt{\frac{10}{9}} + \frac{1}{3}$ $x = +-sqrt(10/9) + 1/3$ ,

or $\approx 1.39$ $~~ 1.39$ and $\approx - 0.72$ $~~-0.72$

Answer 2 · 2017-07-24T05:42:40

$x = \frac{\sqrt{10} + 1}{3}$ $x=(sqrt10+1)/3$
$x = \frac{- (\sqrt{10}) + 1}{3}$ $x=(-(sqrt10)+1)/3$

Explanation:

Given -

$3 x^{2} - 2 x - 3 = 0$ $3x^2-2x-3=0$

Take the constant to the right

$3 x^{2} - 2 x = 3$ $3x^2-2x=3$

Divide both sides by the coefficient of $x^{2}$ $x^2$

$\frac{3 x^{2}}{3} - \frac{2}{3} x = \frac{3}{3}$ $(3x^2)/3-2/3x=3/3$

$x^{2} - \frac{2}{3} x = 1$ $x^2-2/3x=1$

Take half of the coefficient of $x$ $x$ and square it

Half the coefficient of $x = - \frac{2}{3} \div 2 = - \frac{2}{3} \times \frac{1}{2} = - \frac{2}{6}$ $x=-2/3-:2=-2/3xx1/2=-2/6$

Square of $- \frac{2}{6} = \frac{4}{36} = \frac{1}{9}$ $-2/6=4/36=1/9$

Add $\frac{1}{9}$ $1/9$ to both the sides

$x^{2} - \frac{2}{3} x + \frac{1}{9} = 1 + \frac{1}{9} = \frac{9 + 1}{9} = \frac{10}{9}$ $x^2-2/3x+1/9=1+1/9=(9+1)/9=10/9$

${(x - \frac{1}{3})}^{2} = \frac{10}{9}$ $(x-1/3)^2=10/9$

Taking square root on both sides

$x - \frac{1}{3} = \pm \sqrt{\frac{10}{9}} = \pm \frac{\sqrt{10}}{3}$ $x-1/3=+-sqrt (10/9)=+-sqrt10/3$
$x = \pm \frac{\sqrt{10}}{3} + \frac{1}{3} = \pm \frac{\sqrt{10} + 1}{3}$ $x=+-sqrt10/3+1/3= +-(sqrt10+1)/3$

$x = \frac{\sqrt{10} + 1}{3}$ $x=(sqrt10+1)/3$
$x = \frac{- (\sqrt{10}) + 1}{3}$ $x=(-(sqrt10)+1)/3$

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