what is the value of "x" when y=0:
8x^2-x-5=y
eight times x squared minus x minus five equals y
Use the quadratic equation:
x=(1+-sqrt(1+160))/16
check that, I did it in my head.
do you know the anwser to the question above this one
To find the value of "x" when y=0, we need to solve the quadratic equation 8x^2 - x - 5 = 0. This equation can be solved using factoring, completing the square, or the quadratic formula.
Let's use the quadratic formula to find the value of "x":
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 8, b = -1, and c = -5.
Substituting these values into the quadratic formula, we have:
x = (-(-1) ± √((-1)^2 - 4(8)(-5))) / (2(8))
Simplifying further:
x = (1 ± √(1 + 160)) / 16
x = (1 ± √161) / 16
Thus, the two possible solutions for "x" are:
1) x = (1 + √161) / 16
2) x = (1 - √161) / 16