Help with functions

[CowboyVittles]

I'm having trouble understanding the answer to this.

f(x) = 2x^2 + 3x - 4
Find f(1+ [sqrt 2])

The book's answer key says that the answer is: 5 + 7(sqrt 2)
However, when i try to work the problem out, i end up getting: 5 + 3(sqrt 2)

Could somebody please explain to me how to get the correct answer?

 

[Mrspi]

I'm having trouble understanding the answer to this.

f(x) = 2x^2 + 3x - 4
Find f(1+ [sqrt 2])

The book's answer key says that the answer is: 5 + 7(sqrt 2)
However, when i try to work the problem out, i end up getting: 5 + 3(sqrt 2)

Could somebody please explain to me how to get the correct answer?

Substitute (1 + sqrt 2) for x:

f(1 + sqrt 2) = 2(1 + sqrt 2)² + 3(1 + sqrt 2) - 4

And do the arithmetic.

Why don't you show us your steps for doing the arithmetic...and we'll check your work. I got the "book's answer" when I did the problem.

 

[CowboyVittles]

Here's my work
2(1 + [sqrt 2])^2 + 3(1 + [sqrt 2]) - 4=
2(1 + 2) + 3 + (3[sqrt2]) - 4 =
2 + 4 + 3 + (3[sqrt 2]) - 4 =
5 + (3[sqrt2])

This was in a calculus textbook, by the way. But i figured that this could probably pass in the intermediate/advanced algebra section. I thought looking into calculus some over the summer could be something productive, so i bought a book. I am jumping into calculus a bit early though tbh :/

 

[Mrspi]

Here's my work
2(1 + [sqrt 2])^2 + 3(1 + [sqrt 2]) - 4=
2(1 + 2) + 3 + (3[sqrt2]) - 4 =
2 + 4 + 3 + (3[sqrt 2]) - 4 =
5 + (3[sqrt2])

This was in a calculus textbook, by the way. But i figured that this could probably pass in the intermediate/advanced algebra section. I thought looking into calculus some over the summer could be something productive, so i bought a book. I am jumping into calculus a bit early though tbh :/

Your major mistake is in what you did for (1 + sqrt 2)²

That's NOT 1 + 2.

(1 + sqrt 2)² = (1 + sqrt 2)*(1 + sqrt 2)

Multiply each term in the first binomial by each term in the second:

1*1 + 1*sqrt 2 + (sqrt 2)*1 + (sqrt 2)*(sqrt 2)
1 + sqrt 2 + sqrt 2 + 2
3 + 2 sqrt 2

Now .....substitute that for (1 + sqrt 2)²:

2(3 + 2 sqrt 2) + 3 + 3 sqrt 2 - 4
6 + 4 sqrt 2 + 3 + 3 sqrt 2 - 4
(6 + 3 - 4) + [4 sqrt 2 + 3 sqrt 2]
5 + 7 sqrt 2........the "book answer"