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

2 Answers
Jul 24, 2017

Answer:

#x = +-sqrt(10/9) + 1/3#, or #~~ 1.39# and #~~-0.72#

Explanation:

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

#3x^2-2x-3 =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#. After that, we need to take the middle term, #-2/3# and "do some stuff with it" (you'll see in a minute)

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

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

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

#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#, and then immediately subtract #500#, the final result is #0#.

So, if we add #color(red)(1/9)#, and then subtract #color(red)(1/9)#, then technically we haven't changed anything

#x^2-2/3x-1 + color(red)(1/9) + color(red)(-1/9)=0#

Let's re-order this:

#color(green)(x^2-2/3x+1/9) color(blue)(-1 -1/9)=0#

#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.

#color(green)(x^2-2/3x+1/9) =color(green)((x-1/3)^2)#

Let's simplify this: #color(blue)(-1 -1/9)# equals #color(blue)(-10/9)#

So, now we have #color(green)((x-1/3)^2)color(blue)(-10/9)=0#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *#

Now, let's solve this puppy!

#(x-1/3)^2-10/9=0#

add #10/9# to both sides

#(x-1/3)^2 = 10/9#

take a square root of both sides

#sqrt((x-1/3)^2) = +-sqrt(10/9)#

#x-1/3 = +-sqrt(10/9)#

add #1/3# to both sides

#x = +-sqrt(10/9) + 1/3#,

or #~~ 1.39# and #~~-0.72#

Jul 24, 2017

Answer:

#x=(sqrt10+1)/3#
#x=(-(sqrt10)+1)/3#

Explanation:

Given -

#3x^2-2x-3=0#

Take the constant to the right

#3x^2-2x=3#

Divide both sides by the coefficient of #x^2#

#(3x^2)/3-2/3x=3/3#

#x^2-2/3x=1#

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

Half the coefficient of #x=-2/3-:2=-2/3xx1/2=-2/6#

Square of #-2/6=4/36=1/9#

Add #1/9 # to both the sides

#x^2-2/3x+1/9=1+1/9=(9+1)/9=10/9#

#(x-1/3)^2=10/9#

Taking square root on both sides

#x-1/3=+-sqrt (10/9)=+-sqrt10/3#
#x=+-sqrt10/3+1/3= +-(sqrt10+1)/3#

#x=(sqrt10+1)/3#
#x=(-(sqrt10)+1)/3#