SOLUTION: Complete the square for each quadratic equation. A. f(x) = 3x^2 + 6x + 2 B. g(x) = -2x^2 - 12x - 13

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Complete the square for each quadratic equation. A. f(x) = 3x^2 + 6x + 2 B. g(x) = -2x^2 - 12x - 13      Log On


   

Question 1208115: Complete the square for each quadratic equation.


A. f(x) = 3x^2 + 6x + 2

B. g(x) = -2x^2 - 12x - 13
Found 4 solutions by ikleyn, josgarithmetic, timofer, math_tutor2020:
Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

There are no quadratic equations in this post.

There are quadratic functions there.

Or quadratic expressions.

But not quadratic equations.


A person who created this assignment, deserves low mark for incorrect using of mathematical terms.



Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Doing B. only

-2%28x%5E2%2B6x%29-13
-2%28x%5E2%2B6x%2B9-9%29-13 because needed term for completing the square is (6/2)^2=9
-2%28%28x%5E2%2B6x%2B9%29-9%29-13
-2%28%28x%2B3%29%5E2-9%29-13
-2%28x%2B3%29%5E2%2B18-13
highlight%28-2%28x%2B3%29%5E2%2B5%29

Answer by timofer(104) About Me  (Show Source):
You can put this solution on YOUR website!
only doing A.

f%28x%29=3x%5E2%2B6x%2B2
Better to try factor first.
3%28x%5E2%2B2x%29%2B2
If you want to complete the square then you want to use %282%2F2%29%5E2
or 1. You use this for x%5E2%2B2x only.
3%28x%5E2%2B2x%2B1-1%29%2B2
3%28x%2B1%29%5E2-3%2A1%2B2
3%28x%2B1%29%5E2-1----------the answer

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answers:
A. f(x) = 3(x+1)^2 - 1
B. g(x) = -2(x+3)^2 + 5

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Work Shown for Part A

f(x) = 3x^2 + 6x + 2
f(x) = 3(x^2 + 2x) + 2
f(x) = 3(x^2 + 2x + 0) + 2
f(x) = 3(x^2 + 2x + 1-1) + 2 ...... see note below.
f(x) = 3((x^2+2x+1) - 1) + 2
f(x) = 3((x+1)^2 - 1) + 2
f(x) = 3(x+1)^2 + 3(-1) + 2
f(x) = 3(x+1)^2 - 3 + 2
f(x) = 3(x+1)^2 - 1 is the final answer to part A

Note: On the steps marked in blue, I took half of the x coefficient and squared it.
(2/2)^2 = 1^2 = 1
Why do we do this "take half and square it" operation?
See this page
https://www.mathsisfun.com/algebra/completing-square.html

Another example:
y = 2x^2 + 12x + 5
y = 2(x^2 + 6x) + 5
y = 2(x^2 + 6x + 0) + 5
y = 2(x^2 + 6x + 9-9) + 5 ..... half of 6 is 3, which squares to 9
y = 2((x^2+6x+9)-9)+5
y = 2((x+3)^2-9)+5
y = 2(x+3)^2+2(-9)+5
y = 2(x+3)^2-18+5
y = 2(x+3)^2-13

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Explanation for Part B

We could use the same process as part A. However, I'll use a different approach.
Recall that y = ax^2+bx+c is standard form while y = a(x-h)^2+k is vertex form
(h,k) is the location of the vertex.

h = -b/(2a) is the formula to find the x coordinate of the vertex.

We have y = -2x^2 - 12x - 13 where the coefficients are: a = -2, b = -12, c = -13
So,
h = -b/(2a)
h = -(-12)/(2*(-2))
h = -3
Plug this value into the equation to find its paired y value.
y = -2x^2 - 12x - 13
y = -2(-3)^2 - 12(-3) - 13
y = -2(9) - 12(-3) - 13
y = -18 + 36 - 13
y = 18 - 13
y = 5
This is the y coordinate of the vertex, so we have k = 5.

h = -3 and k = 5 pair up to give the vertex (h,k) = (-3,5)
a = -2 from earlier

We go from
y = a(x-h)^2+k
to
y = -2(x-(-3))^2 + 5
then that simplifies into
y = -2(x+3)^2 + 5 which is the answer to part B.

The CompleteSquare function in GeoGebra is useful to verify the answers.