Algebra 2
posted by Ana on .
How would you find the x intercepts of 2x^2+17=0. The original equation is 2(x3)^21.

To have xintercepts you must have a function,
I bet your equation was
y = 2(x3)^2  1 or f(x) = 2(x3)^2 1
then
y = 2(x^2  6x + 9)  1
y = 2x^2  12x + 17 , your expansion was incorrect
so at the xintercept, y = 0 and
2x^2  12x + 17 = 0
x = (12 ± √8)/4
= (12 ± 2√2)/4
x = (6+√2)/2 or x = (6  √2)/2 
thanks so much!