Select the approximate values of x that are solutions to f(x) = 0, where
f(x) = -9x2 + 9x + 5.
(a){–1.80, 1.80}
(b){-9, 9}
(c){–1.00, –0.56}
(d){–0.40, 1.40}
it's 9x^2 not 9x2
so -9x^2 + 9x + 5 = 0
9x^2 - 9x - 5 = 0
Don't see any obvious facors, so let's use the formula
x (9 ± √261)/18
= appr 1.4 or -.4
looks like d)
To find the approximate values of x that are solutions to f(x) = 0, we can use the quadratic formula. The quadratic formula states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the quadratic equation is -9x^2 + 9x + 5 = 0, which means a = -9, b = 9, and c = 5.
Using the quadratic formula, we have:
x = (-9 ± √(9^2 - 4(-9)(5))) / (2(-9))
Simplifying further, we have:
x = (-9 ± √(81 + 180)) / (-18)
x = (-9 ± √(261)) / (-18)
Now, we need to evaluate the two possible solutions:
Solution 1: x = (-9 + √(261)) / (-18)
Solution 2: x = (-9 - √(261)) / (-18)
Now, let's calculate the approximate values for these solutions:
Solution 1:
x = (-9 + √(261)) / (-18)
Using a calculator, we find:
x ≈ -0.405
Solution 2:
x = (-9 - √(261)) / (-18)
Using a calculator, we find:
x ≈ 1.405
So, the approximate values of x that are solutions to f(x) = 0 are:
(d) { -0.40, 1.40 }