how do you get the x inercept of a equation like 7/(4x^2-21x-49)? I know you put it in y=mx+b but i don't know how to solve for x in this
To find the x-intercept of an equation, you need to set the value of y to zero and solve for x. In this case, the given equation is: y = 7/(4x^2 - 21x - 49).
To start, set y = 0:
0 = 7/(4x^2 - 21x - 49)
Since the numerator is not equal to zero, the equation can only be satisfied if the denominator is equal to zero. So, we have:
4x^2 - 21x - 49 = 0
To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula to find the value of x:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the equation to standard quadratic form ax^2 + bx + c = 0, we have:
a = 4, b = -21, c = -49
Substituting these values into the quadratic formula, we get:
x = (-(-21) ± √((-21)^2 - 4 * 4 * -49)) / (2 * 4)
Simplifying further:
x = (21 ± √(441 + 784)) / 8
x = (21 ± √1225) / 8
Since we are looking for the x-intercept, we only need the real solutions. Evaluating the square root:
x = (21 ± 35) / 8
This gives us two possible solutions:
x1 = (21 + 35) / 8 = 56 / 8 = 7/1 = 7
x2 = (21 - 35) / 8 = -14 / 8 = -7/4
Therefore, the x-intercepts of the equation 7/(4x^2 - 21x - 49) are x = 7 and x = -7/4.