Question

F(x)=(x+3)^2 -8 how many real solutions?

Answer 1

2 real solutions

Explanation:

`f(x)=(x+3)^2-8\\\\0=(x+3)^2-8\\\\8=(x+3)^2\\\\\pm√(8)=x+3\\\\x=-3\pm√(8)`

Therefore, the function set at
`f(x)=0` has two real solutions (x-intercepts).

You can also check the discriminant of the function by expanding it:

`(x+3)^2-8=x^2+6x+9-8=x^2+6x+1\\\\b^2-4ac=6^2-4(1)(1)=36-4=32 > 0`

Since it's greater than 0, there's two real solutions