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(x-3)^2-1.

- Algebra 2 -
**Reiny**, Friday, December 2, 2011 at 11:12am
To have x-intercepts you must have a function,

I bet your equation was

y = 2(x-3)^2 - 1 or f(x) = 2(x-3)^2 -1

then

y = 2(x^2 - 6x + 9) - 1

y = 2x^2 - 12x + 17 , your expansion was incorrect

so at the x-intercept, 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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