Determine X in the Function y = -x² + 2x + 35 Where f(x) > 0

Quadratic Inequalities with Factored Form

Look at the following function:

y=x2+2x+35 y=-x^2+2x+35

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

Follow each step carefully to understand the complete solution
1

Understand the problem

Look at the following function:

y=x2+2x+35 y=-x^2+2x+35

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

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

2

Step-by-step solution

To solve for when the quadratic function y=x2+2x+35>0 y = -x^2 + 2x + 35 > 0 , we must first find the roots of the function using the quadratic formula. The quadratic function is given as y=x2+2x+35 y = -x^2 + 2x + 35 .

Step 1: Calculate the roots using the quadratic formula. For ax2+bx+c=0 ax^2 + bx + c = 0 , the formula is:

x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here, a=1 a = -1 , b=2 b = 2 , and c=35 c = 35 . Thus, we compute the discriminant:

b24ac=224(1)(35)=4+140=144 b^2 - 4ac = 2^2 - 4(-1)(35) = 4 + 140 = 144

Since the discriminant is positive, there are two distinct real roots.

Step 2: Compute the roots using the quadratic formula:

x=2±1442(1)=2±122 x = \frac{-2 \pm \sqrt{144}}{2(-1)} = \frac{-2 \pm 12}{-2}

Calculating the two roots, we get:

x1=2+122=102=5 x_1 = \frac{-2 + 12}{-2} = \frac{10}{-2} = -5 x2=2122=142=7 x_2 = \frac{-2 - 12}{-2} = \frac{-14}{-2} = 7

Step 3: Determine the intervals where f(x)>0 f(x) > 0 . The roots x=5 x = -5 and x=7 x = 7 partition the number line into intervals. A quadratic function with a negative leading coefficient a a opens downward, meaning it is positive between its roots:

The intervals are:

  • (,5)(-∞, -5)
  • (5,7)(-5, 7)
  • (7,)(7, ∞)

Test the interval between the roots: Choose a point, say x=0 x = 0 , between 5-5 and 77:

f(0)=(0)2+2(0)+35=35>0 f(0) = -(0)^2 + 2(0) + 35 = 35 > 0

This confirms that the function is positive in the interval (5,7)(-5, 7).

Therefore, the solution to the inequality f(x)>0 f(x) > 0 is 5<x<7-5 < x < 7.

The solution to the problem is 5<x<7-5 < x < 7.

3

Final Answer

5<x<7 -5 < x < 7

Key Points to Remember

Essential concepts to master this topic
  • Rule: Find roots first, then determine sign intervals between roots
  • Technique: Use quadratic formula: x=2±122 x = \frac{-2 \pm 12}{-2} gives -5 and 7
  • Check: Test point between roots: f(0)=35>0 f(0) = 35 > 0 confirms interval ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting parabola opens downward with negative coefficient
    Don't assume function is positive outside the roots like upward parabolas = wrong interval! With coefficient a = -1, the parabola opens downward, making it negative outside roots. Always check the sign of the leading coefficient to determine which interval is positive.

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 if I'm solving an inequality?

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The roots are boundary points where the function changes from positive to negative! They help you identify the intervals to test. Without finding where f(x)=0 f(x) = 0 , you can't determine where f(x)>0 f(x) > 0 .

How do I know which interval to choose?

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After finding roots, test a point from each interval! Pick any number between the roots and substitute it into the function. If the result is positive, that's your answer interval.

What if the discriminant is negative?

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If b24ac<0 b^2 - 4ac < 0 , there are no real roots! The parabola doesn't cross the x-axis. Check if a is positive (always positive) or negative (always negative).

Why is the answer -5 < x < 7 and not -5 ≤ x ≤ 7?

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We need f(x)>0 f(x) > 0 , which means strictly greater than zero. At the roots x = -5 and x = 7, the function equals zero, so we use < instead of ≤.

Can I solve this by graphing instead?

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Absolutely! Graph y=x2+2x+35 y = -x^2 + 2x + 35 and look for where the curve is above the x-axis. The x-values where y > 0 give you the same answer: between -5 and 7.

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