SOLUTION: Hello! I need some help with this equation. Solve y = 2x^2 + 1 when x = -2. I got y= -7, but am unsure if it is the correct answer. Any assistance would be greatly appreciat

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Question 1174347: Hello! I need some help with this equation.
Solve y = 2x^2 + 1 when x = -2.
I got y= -7, but am unsure if it is the correct answer. Any assistance would be greatly appreciated!
Found 2 solutions by math_tutor2020, mccravyedwin:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

This might be overkill, but think of x as (x)

So we're dealing with y = 2(x)^2 + 1

Replace every copy of x with -2 and evaluate to get

y = 2(x)^2 + 1
y = 2(-2)^2 + 1
y = 2(-2)*(-2) + 1
y = 2(4) + 1
y = 8 + 1
y = 9

I have a feeling your steps probably looked like this
y = 2x^2 + 1
y = 2(-2)^2 + 1
y = 2(-4) + 1
y = -8 + 1
y = -7
Which is not correct because (-2)^2 is equal to positive 4, and not negative 4.
Think of (-2)^2 as (-2)*(-2) = 4. The two negatives multiply to a positive.
In short, we're squaring all of the value and that includes the negative.

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Answer: y = 9
Answer by mccravyedwin(407)   (Show Source): You can put this solution on YOUR website!

Substitute (-2) in place of x in 

y = 2x^2 + 1

We get

y = 2(-2)^2 + 1

Substitute (4) in place of (-2)^2 because -2 times itself gives +4

y = 2(4) + 1

Substitute 8 in place of 2(4) because 2 times 4 is 8

y = 8 + 1

Substitute 9 in place of 8 + 1 because 8 plus 1 is 9

y = 9.

Oh, oh, you got -7.  See if you can find out where you went wrong.

Edwin
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