Finding Positive Domains: Solve (x-1/2)(-x+7/2) > 0

Quadratic Inequalities with Sign Analysis

Find the positive and negative domains of the function below:

y=(x12)(x+312) y=\left(x-\frac{1}{2}\right)\left(-x+3\frac{1}{2}\right)

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

f(x)>0 f(x) > 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 below:

y=(x12)(x+312) y=\left(x-\frac{1}{2}\right)\left(-x+3\frac{1}{2}\right)

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

f(x)>0 f(x) > 0

2

Step-by-step solution

To solve this problem, we'll determine when the product (x12)(x+312) (x - \frac{1}{2})(-x + 3\frac{1}{2}) is positive. This involves finding the roots of the equation and testing the intervals between these roots:

Step 1: **Determine the roots of the factors.**
- The first factor x12=0 x - \frac{1}{2} = 0 gives the root x=12 x = \frac{1}{2} .
- The second factor x+312=0 -x + 3\frac{1}{2} = 0 gives the root x=312 x = 3\frac{1}{2} .

Step 2: **Identify intervals based on these roots.**
- The roots divide the x x -axis into three intervals: x<12 x < \frac{1}{2} , 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} , and x>312 x > 3\frac{1}{2} .

Step 3: **Analyze the sign of the function in each interval.**
- For x<12 x < \frac{1}{2} :
- x12<0 x - \frac{1}{2} < 0 and x+312>0 -x + 3\frac{1}{2} > 0 , so the product is negative.
- For 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} :
- Both x12>0 x - \frac{1}{2} > 0 and x+312>0 -x + 3\frac{1}{2} > 0 , so the product is positive.
- For x>312 x > 3\frac{1}{2} :
- x12>0 x - \frac{1}{2} > 0 and x+312<0 -x + 3\frac{1}{2} < 0 , so the product is negative.

Therefore, the intervals where y>0 y > 0 are 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} .

This matches the given correct answer choice: 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} .

3

Final Answer

12<x<312 \frac{1}{2} < x < 3\frac{1}{2}

Key Points to Remember

Essential concepts to master this topic
  • Roots: Find where each factor equals zero to determine critical points
  • Technique: Test intervals between roots: x=0 x = 0 gives (12)(312)<0 (-\frac{1}{2})(3\frac{1}{2}) < 0
  • Check: Substitute x=1 x = 1 : (12)(212)=114>0 (\frac{1}{2})(2\frac{1}{2}) = 1\frac{1}{4} > 0

Common Mistakes

Avoid these frequent errors
  • Testing the wrong side of inequality sign
    Don't solve for when the product is negative when the problem asks for positive = opposite answer! Students often confuse > 0 with < 0 and get intervals where the function is negative instead. Always double-check whether you need positive or negative values.

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 find the roots first?

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The roots are where the function equals zero, creating boundaries between positive and negative regions. They're like dividing lines that separate where your inequality is true from where it's false.

How do I know which interval to test?

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Pick any number inside each interval and substitute it into the original expression. If the result matches your inequality (> 0 or < 0), that entire interval is part of your solution!

What if I get confused about the signs?

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Make a sign chart! List each factor, mark + or - in each interval, then multiply the signs. Remember: positive × positive = positive and negative × negative = positive.

Do the boundary points count in my answer?

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For strict inequalities (> or <), the boundary points where the function equals zero are not included. Only use ≥ or ≤ when the problem specifically allows equal to zero.

Can I expand the expression instead?

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You could, but it's much harder! Factored form makes it easy to find roots and test intervals. Keep it factored - it's the most efficient method for solving inequalities.

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