Given: f(x) = X^2-x
evaluate: f(sqrt2 + 3)
I have worked out this and am stuck:
(sqrt2+3)^2 - (sqrt2+3)
(sqrt2+3)*(sqrt2+3)- (sqrt2 +3)
(sqrt2)^2+ 3sqrt2 +3sqrt2+ 9- (sqrt2+3)
2+6sqrt2 +9-sqrt2-3
and am stuck or wrong here. Ans= SQRT2+6sqrt2 all under radical
(sqrt2+3)^2 - (sqrt2+3)
let a = sqrt 2 so a^2 = 2
(a+3)^2 - (a+3)
a^2 + 6 a + 9 - a - 3
a^2 + 5 a + 6
2 + 5 sqrt 2 + 6
8 + 5 sqrt 2
THANKS!!
To evaluate the expression f(sqrt2 + 3), you need to substitute sqrt2 + 3 into the function f(x).
Given: f(x) = x^2 - x
Replacing x with sqrt2 + 3, we have:
f(sqrt2 + 3) = (sqrt2 + 3)^2 - (sqrt2 + 3)
Now let's simplify this step by step:
1. Expand (sqrt2 + 3)^2 using the formula (a + b)^2 = a^2 + 2ab + b^2:
= (sqrt2)^2 + 2(sqrt2)(3) + 3^2 - (sqrt2 + 3)
= 2 + 2(sqrt2)(3) + 9 - (sqrt2 + 3)
= 2 + 6sqrt2 + 9 - sqrt2 - 3
2. Combine like terms:
= 2 - 3 + 6sqrt2 - sqrt2 + 9
= 8 + 5sqrt2
Therefore, the simplified expression for f(sqrt2 + 3) is 8 + 5sqrt2.