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f(x) = 2x^2 - x + 3

f ( -a) =

i got this for an answer -2a^2 + a + 3

but the book says im wrong and the (-2a^2) has to be a positive.... why is that ?

2007-12-23 12:22:05 · 15 answers · asked by Anonymous in Science & Mathematics Mathematics

15 answers

The reason being is that it is the a that is being squared...not the -2a.

So, a new equation would look like this:

2((-a)^2) - (-a) + 3

The 2((-a)^2) has a negative a times a negative which is positive a squared...

2a^2+a+3 is the final answer

2007-12-23 12:26:54 · answer #1 · answered by Ninja Noramros 2 · 0 0

if I am reading your original problem correctly it is because....

the X is squared. The parenthesis indicates that the -a includes the negative sign in being squared.

So 2(-a)^2 = a positive 2a^2 b/c any negative squared is a positive. Does that make sense?

2007-12-23 12:27:59 · answer #2 · answered by USMC_Wifey 3 · 0 0

Because when you square the x, you make the negative sign positive. For example, -3^2 is 9, not -9. So, -a^2 is a^2

2007-12-23 12:24:54 · answer #3 · answered by Anonymous 2 · 0 0

I think you're right. Remember that the people who write text books do make mistakes. I can remeber a few times where I kept re-doing a problem because the book's answer was different from mine. Then the next day in class, the teacher said, "Oh ya, the book is wrong."
Hopefully someone else will respond soon and maybe teah both you and me someting we didn't know or forgot.

2007-12-23 12:28:51 · answer #4 · answered by Rich 4 · 0 0

You are substituting -a for x.
When a gets squared it is minus times minus which is plus.
Answer should be 2a^2 +a +3

2007-12-23 12:27:02 · answer #5 · answered by ignoramus 7 · 0 0

The final squaring is beside the point. the only way (...)^2 could be 0 is for (...) to be 0. so sparkling up 5z^2 + 6z + 12 = 0 The quadratic formulation z = (-b + or - sqrt(b^2 - 4ac))/2a with a = 5, b = 6, c = 12 will provide a unfavourable cost for b^2 - 4ac, meaning you get complicated roots. Do it your self, yet i think of you get z = -0.6 + 0.2*sqrt(fifty one)*i or z = -0.6 - 0.2*sqrt(fifty one)*i

2016-11-24 21:40:00 · answer #6 · answered by ? 4 · 0 0

f(-a) = 2(-a)^2 -(-a) + 3
f(-a) = 2(a^2) + a + 3 = 2a^2 = a + 3

because (-a)(-a) = a^2

2007-12-23 12:25:55 · answer #7 · answered by MartinWeiss 6 · 0 0

2(-a)^2 - (-a) + 3

When you square the variable (-a) it is essentially the same as multiplying two negative numbers. Anytime you multiply an even amount of negative numbers or variables your result will be positive.

2007-12-23 12:31:42 · answer #8 · answered by ninjagirl1001 2 · 0 0

because the (-a) is squared before it is multiplied by 2. Any number that is a product of two negative numbers is a positive number.

2007-12-23 12:27:58 · answer #9 · answered by sparky 1 · 0 0

[03]
Yes,your book is right.Because 2(-a)^2=2a^2 and not -2a^2
Please remember (-a)^2=-a*-a=a^2

2007-12-23 12:27:12 · answer #10 · answered by alpha 7 · 0 0

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