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

Quadratic Inequalities with Factored Form

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(x)>0 f\left(x\right) > 0

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Step-by-step written solution

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1

Understand the problem

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(x)>0 f\left(x\right) > 0

2

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}

3

Final Answer

x>12 x > \frac{1}{2} or x<612 x < -6\frac{1}{2}

Key Points to Remember

Essential concepts to master this topic
  • Zeros: Set each factor equal to zero to find roots
  • Sign Analysis: Test intervals between roots: x = -7 gives (-)(+) = negative
  • Check: Verify positive regions match parabola opening upward ✓

Common Mistakes

Avoid these frequent errors
  • Solving the inequality like an equation
    Don't just find where (x - 1/2)(x + 6.5) = 0 and stop = only gives boundary points! This misses the actual regions where the function is positive or negative. Always analyze the sign in each interval between the zeros.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

AAABBBCCCX

FAQ

Everything you need to know about this question

Why do I need to check the sign in each interval?

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The quadratic function changes from positive to negative (or vice versa) at each zero. By testing a point in each interval, you determine whether the function is above or below the x-axis in that region.

How do I remember which intervals are positive?

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Since this is a parabola opening upward (positive leading coefficient), it's positive on the outside intervals and negative on the inside interval between the zeros.

What if I get confused with the mixed numbers?

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Convert mixed numbers to improper fractions first: 612=132 6\frac{1}{2} = \frac{13}{2} . This makes the calculations cleaner and less error-prone.

Can I just look at the graph instead?

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Absolutely! Graphing helps visualize where the parabola is above the x-axis (positive). But algebraic analysis ensures you get the exact boundary points.

Why does the answer use 'or' instead of 'and'?

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The function is positive in two separate regions that don't connect. Use 'or' when the solution includes disconnected intervals, and 'and' only when values must satisfy multiple conditions simultaneously.

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