Solve (x-1/2)(x+6.5): Finding Positive Domain of Quadratic Function

Question

Find the positive and negative domains of the function:

y=(x12)(x+612) y=\left(x-\frac{1}{2}\right)\left(x+6\frac{1}{2}\right)

Determine for which values of x x the following is true:

f\left(x\right) > 0

Step-by-Step Solution

To find when the function y=(x12)(x+612) y = \left(x - \frac{1}{2}\right)\left(x + 6\frac{1}{2}\right) is positive, we proceed as follows:

First, identify the roots of the expression by solving x12=0 x - \frac{1}{2} = 0 and x+612=0 x + 6\frac{1}{2} = 0 . These calculations give us the roots x=12 x = \frac{1}{2} and x=612 x = -6\frac{1}{2} , or x=132 x = -\frac{13}{2} .

Next, determine the sign of the product (x12)(x+132) (x - \frac{1}{2})(x + \frac{13}{2}) over the intervals defined by these roots:

  • Interval 1: x<132 x < -\frac{13}{2} . In this region, both x12 x - \frac{1}{2} and x+132 x + \frac{13}{2} are negative, so their product is positive.
  • Interval 2: 132<x<12 -\frac{13}{2} < x < \frac{1}{2} . In this region, x+132 x + \frac{13}{2} is positive and x12 x - \frac{1}{2} is negative, so their product is negative.
  • Interval 3: x>12 x > \frac{1}{2} . In this region, both x12 x - \frac{1}{2} and x+132 x + \frac{13}{2} are positive, so their product is positive.

Therefore, the function is positive for x<612 x < -6\frac{1}{2} and x>12 x > \frac{1}{2} .

Thus, the solution is:
x>12 x > \frac{1}{2} or x<612 x < -6\frac{1}{2}

Answer

x > \frac{1}{2} or x < -6\frac{1}{2}