Posted by Ana on Friday, December 2, 2011 at 11:04am.
How would you find the x intercepts of 2x^2+17=0. The original equation is 2(x3)^21.

Algebra 2  Reiny, Friday, December 2, 2011 at 11:12am
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

Algebra 2  Ana, Friday, December 2, 2011 at 12:06pm
thanks so much!
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