Asked by Ana
How would you find the x intercepts of 2x^2+17=0. The original equation is 2(x-3)^2-1.
Answers
Answered by
Reiny
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
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
Answered by
Ana
thanks so much!