# How do you solve using the completing the square method #3x^2-67=6x+55#?

##### 2 Answers

#### Answer:

#### Explanation:

#"rearrange into standard form"#

#•color(white)(x)ax^2+bx+c=0#

#rArr3x^2-6x-122=0larrcolor(blue)"in standard form"#

#"to "color(blue)"complete the square"#

#• " coefficient of "x^2" term must be 1"#

#rArr3(x^2-2x-122/3)=0#

#• " add/subtract "(1/2"coefficient of x-term")^2" to"#

#x^2-2x#

#3(x^2+2(-1)xcolor(red)(+1)color(red)(-1)-122/3)=0#

#rArr3(x-1)^2+3(-1-122/3)=0#

#rArr3(x-1)^2-125=0#

#rArr3(x-1)^2=125#

#rArr(x-1)^2=125/3#

#color(blue)"take square root of both sides"#

#rArrx-1=+-sqrt(125/3)larrcolor(blue)"note plus or minus"#

#rArrx=1+-(5sqrt5)/sqrt3#

#rArrx=1+-(5sqrt15)/3larrcolor(blue)"rationalise denominator"#

#rArrx~~-5.45" or "x~~7.45" to 2 dec. places"#

#### Answer:

#### Explanation:

Given:

By manipulating this we obtain

Set this as

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Consider the perfect square format of

Our objective is to manipulate **close as we can** to the perfect square format. This process introduces an error. This error is compensated for by including an 'adjustment' that takes the final result back to the value it should be. This correction I will show as

Write as

Now we will

We have halved the

We need to set

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Add

Divide both sides by 3

Square root both sides

Mathematically it is considered that to have a root in the denominator is not good practice. So lets get rid of it.

Multiply the roots by 1 but in the form of