# Algebra 2

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How would you find the x intercepts of 2x^2+17=0. The original equation is 2(x-3)^2-1.

• Algebra 2 -

To have x-intercepts you must have a function,
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 -

thanks so much!

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