How would you find the x intercepts of 2x^2+17=0. The original equation is 2(x-3)^2-1.
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
thanks so much!
To find the x-intercepts of the quadratic equation 2x^2 + 17 = 0, you can use the quadratic formula. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the given equation 2x^2 + 17 = 0 to the standard quadratic equation ax^2 + bx + c = 0, we see that a = 2, b = 0, and c = 17. Plugging these values into the quadratic formula, we get:
x = (0 ± √(0^2 - 4*2*17)) / (2*2)
x = ±√(-136) / 4
Since the expression inside the square root is negative, there are no real solutions to this quadratic equation. Therefore, there are no x-intercepts for the equation 2x^2 + 17 = 0.
For the second equation, 2(x-3)^2 - 1, it is an equation of a parabola that has been shifted horizontally 3 units to the right and downward 1 unit.
By setting the equation equal to zero, we can find the x-intercepts. So, we have:
2(x-3)^2 - 1 = 0
Expanding the equation, we get:
2(x^2 - 6x + 9) - 1 = 0
2x^2 - 12x + 18 - 1 = 0
2x^2 - 12x + 17 = 0
Now, you can use the quadratic formula to find the x-intercepts. Comparing this equation to the standard form ax^2 + bx + c = 0, we have a = 2, b = -12, and c = 17.
Plugging these values into the quadratic formula:
x = (-(-12) ± √((-12)^2 - 4*2*17)) / (2*2)
x = (12 ± √(144 - 136)) / 4
x = (12 ± √8) / 4
Simplifying further:
x = (12 ± 2√2) / 4
x = 3 ± (1/√2)
Therefore, the x-intercepts of the equation 2(x-3)^2 - 1 are approximately 3 + (1/√2) and 3 - (1/√2).