Let f(x)= x^2-5x+5?

Question

Let f(x)= x^2-5x+5

f(1+ root 3)


f(1+root 3) =


Can someone please explain this equation to me? I'll give best answer :)

Answer 1

I assume you mean: 1 + the square root of 3, correct?

First: replace (1+V`3) with the x-variable....

f(x) = x^2 - 5x + 5

f(1 + V`3) = (1 + V`3)^2 - 5(1 + V`3) + 5

Sec: solve the function...

f(1 + V`3) = (1 + V`3)(1 + V`3) - 5(1 + V`3) + 5

= 1(1)+1(V`3)+(V`3)(1)+(V`3)(V`3) - 5(1) - 5(V`3)+5

= 1 + V`3 + V`3 + (V`9) - 5 - 5V`3 + 5

= 1 + V`3 + V`3 + 3 - 5 - 5V`3 + 5

= 1 + 2V`3 + 3 - 5 - 5V`3 + 5

Third: combine "like" terms...

= 1 + 3 - 5 + 5 + 2V`3 - 5V`3

= 4 - 3V`3

Answer 2

f(x)= x^2-5x+5 is the definition of a function. A function is like a machine with ins and outs: in come a value for x, out go the result of the expression with that x, called here f(x).

What is required is to calculate f(1 + sqrt(3)); in other words, in the expression x^2 - 5x + 5, substitute every x by 1 + sqrt(3), and simplify the resulting numeric expression.

Just to give a hand in the calculations: (1 + sqrt(3) )^2 = 1 + 2*sqrt(3) + 3 = 4 + 2 * sqrt(3).

Answer 3

f(x) = x^2 - 5x + 5

And you want to solve

f(1 + sqrt(3)) and
f(1 - sqrt(3)) {I'm assuming you mistyped}.

First, let's solve f(1 + sqrt(3)). We plug in every occurrence of x with (1 + sqrt(3)).

f[1 + sqrt(3)] = [1 + sqrt(3)]^2 - 5[1 + sqrt(3)] + 5

First, expand the squared part. This is no different from expanding (a + b)^2, which would be a^2 + 2ab + b^2.
That means [1 + sqrt(3)]^2 would be (1^2 + 2sqrt(3) + [sqrt(3]^2), but [sqrt(3)]^2 is equal to 3, and that gives us:

f(1 + sqrt(3)) = (1 + 2sqrt(3) + 3) - 5[1 + sqrt(3)] + 5

Now, we distribute the -5.

1 + 2sqrt(3) + 3 - 5 - 5sqrt(3) + 5

Combine like terms,

4 + 2sqrt(3) - 5sqrt(3)

And again, simplify

4 - 3sqrt(3)

Answer 4

f(1+sqrt(3))=
(1+sqrt(3))^2 -5(1+sqrt(3))+5=
1+3+2sqrt(3)-5-5sqrt(3)+5=
4-3sqrt(3)

Answer 5

i could help you if you told me what went in the place of x. your problem doesn't have any domains inputs?