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Stels [109]
3 years ago
13

What are the solutions to m^2 – 9 = 0? a. m = –9 and m = 0 b. m = –3 and m = 3 c. m = –3 d. m = 9

Mathematics
1 answer:
marysya [2.9K]3 years ago
5 0
2 ways
zero product property
easy way

zero product
factor perfect square
m^2-3^2=0
(m-3)(m+3)=0
set each to zero
m-3=0
x=3

m+3=0
m=-3

m=-3 or 3

easy way
add 9 to both sides
m^2=9
sqrt both sides remember to take postive and negative roots
m=+/-3
m=-3 or 3


B is answer

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Step-by-step explanation:

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but before we do that, we will use the vertex form, since it will make our work easier at the beginning.  

First and foremost, when we plot the vertex and the given point, the vertex is higher up than is the point; that means that this parabola opens upside down, and its vertex form will be

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The absolute value is out in front of the a, so we know that the value of a is positive, but the quadratic itself is negative (upside down) and we will find that math takes care of that negative that needs to be out front.  So we need to solve for a by filling in the x, y, h, and k values from the point and the vertex:  x = -7, y = 0, h = -3, k = 4

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Now that we know a, we can plug it back into the vertex form and then put it into standard form from there.

y=-\frac{1}{4}(x+3)^2+4

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y=-\frac{1}{4}(x^2+6x+9)+4

Simplify by distributing the -1/4 into the parenthesis:

y=-\frac{1}{4}x^2-\frac{3}{2}x-\frac{9}{4}+4

Combine like terms to get

y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{7}{4}

And there you go!

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