SOLUTION: Complete the square for each quadratic equation.
A. f(x) = 3x^2 + 6x + 2
B. g(x) = -2x^2 - 12x - 13
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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) (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) (Show Source): You can put this solution on YOUR website!
Doing B. only
because needed term for completing the square is (6/2)^2=9
Answer by timofer(104) (Show Source): You can put this solution on YOUR website!
only doing A.
Better to try factor first.
If you want to complete the square then you want to use
or 1. You use this for only.
----------the answer
Answer by math_tutor2020(3816) (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.
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