Number 17
To solve this, you simply set y = 0, and solve for x. The right-most x will be a bigger value than the leftmost x
ok i got to x^2-6x+6 ...where do i got from here?
x^2 - 6x + 6 = 0
i plugged zero in for y, foiled on one side, and subtracted 3
Did you figure it out yet?
nope, just got the wrong answer
o wait, do i use the quadratic formula?
Try (4.732, 0)
its just 4.732 because its only asking for the x coordinate
I see
so how did u get that #?
I approximated \[x = \sqrt{3} + 3\]
how did u get that from x^2-6x+6=0?
Well, to be honest, I back tracked to when you had 3 = (x-3)^2 Instead of putting everything on one side, I simply square rooted both sides to get: sqrt{3} = x - 3 Then I added 3 to both sides. Why? Because we wanted to solve for x.
Ok I see
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