# Quadratic Functions, and Completing the Square

#### H.M. Murdock

I had a hard time completing the square of some problems. This is what I did with them, I would appreciate if someone could show me the mistake that im doing and the right process in order to solve these problems.
Also, here there is an article about completing the square for anyone that dosent remeber about it.
http://www.uncwil.edu/courses/mat111hb/Izs/asolve/asolve.html#compsqr [Broken]

First problem
Complete the square on y = 3xsquare - x + 1 to write it in the form y = a (x - h)square +k

y = 3xsquare - x +1

3xsquare/3 and - x/3 =

3 (xsquare - 1/3x) +1

(1/3)square = 1/9

3 (xsquare -1/3x + 1/9) + 1 - 1/3

The answer is y = 3 (x + 1/3) square 2/3

But the answer on my book is y = 3 (x - 1/6) square + 11/12
I d like to know what was wrong there.

Second problem
Complete the square on y = xsquare - 6x + 10 to write it in the form y = a (x -h)square +k

y = xsquare - 6x + 10

(xsquare -6x) + 10

(-6/2)square = 9

(xsquare - 6x + 9) 10 - 9

The answer is (x + 3)square + 1

However on my book the aswer is y = (x - 3)square + 1
I dont really know where did that "-3" came from.

Third problem
-What are the x- and y- intercepts for f(x) = 2xsquare + x - 6 ?

This quadratic funtions is in the form " axsquare + bx + c " so I used the formula " h = -b/2a "
in order to solve it.

a = 2 and b = 1 on the function

so h = -1/2(2) = -1/4

k = 2(-1/4) square + (-1/4) - 6

k = 1/8 - 1/4 - 6

k= 49/8

I would appreciate if anyone is able to tell me how to solve these 3 problems.
Thanks a lot in advance.

Last edited by a moderator:
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#### happyg1

Hi,
on problem 1, You have to divide ALL of the terms by 3. I did this, and i didn't get what your answer is:
$$\frac{3x^2}{3}-\frac{x}{3}+\frac{1}{3}=0$$
$$x^2-\frac{1}{3}x=-\frac{1}{3}$$
that third term is then:
$$\left(\frac{1}{2}\cdot\frac{1}{3}\right)^2=\frac{1}{36}$$
then add that to both sides:
$$x^2-\frac{1}{3}x+\frac{1}{36}=-\frac{1}{3}+\frac{1}{36}$$
see what you get from there.

Second problem:
you got
$$x^2-6x+9=-1$$
when you factor that you have:
$$(x-3)^2+1$$ (after you add the one back over)

Third problem:
You found the x-value for the minimum, not the x and y intercepts.
For x intercepts, set it equal to zero and solve.
Your y-int. is f(0).
CC

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#### Redbelly98

Staff Emeritus
Science Advisor
Homework Helper
First problem:

as happyg1 said, you don't square (1/3). First you divide by 2, and then square that:

[(1/2)*(1/3)]^2 = (1/6)^2 = 1/36

and so

x^2 - x/3 + 1/36
is a perfect square, equal to
(x - 1/6)^2

#### H.M. Murdock

First problem:

as happyg1 said, you don't square (1/3). First you divide by 2, and then square that:

[(1/2)*(1/3)]^2 = (1/6)^2 = 1/36

and so

x^2 - x/3 + 1/36
is a perfect square, equal to
(x - 1/6)^2
Hi thanks a lot guys, I appreciate it , but where does the 2 of the division come from?

#### Dick

Science Advisor
Homework Helper
(x+a)^2=x^2+2xa+a^2. So if you see x^2+Ax, if it's part of the complete square (x+a)^2, A=2a. Just equate the coefficients of the linear factors.

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