Solve: When is -x² + 2x + 35 Less Than Zero?

Quadratic Inequalities with Sign Analysis

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 determine where f(x)<0 f(x) < 0 for the given quadratic function y=x2+2x+35 y = -x^2 + 2x + 35 , we'll perform the following steps:

  • Step 1: Identify the roots of the function using the quadratic formula.
  • Step 2: Analyze the intervals around the roots to establish where the function is negative.

Step 1: Find the roots using the quadratic formula:

The quadratic formula is given by:

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

For our function y=x2+2x+35 y = -x^2 + 2x + 35 , we have a=1 a = -1 , b=2 b = 2 , and c=35 c = 35 . Substituting into the formula:

x=2±224(1)(35)2(1) x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(35)}}{2(-1)}

x=2±4+1402 x = \frac{-2 \pm \sqrt{4 + 140}}{-2}

x=2±1442 x = \frac{-2 \pm \sqrt{144}}{-2}

x=2±122 x = \frac{-2 \pm 12}{-2}

This gives two roots:

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

Step 2: Analyze the intervals created by the roots:

The roots divide the number line into the intervals x<7 x < -7 , 7<x<5-7 < x < -5, and x>5 x > -5 .

Since the parabola y=x2+2x+35 y = -x^2 + 2x + 35 opens downwards, it will be less than 0 outside the region between the roots. Therefore, the intervals where y<0 y < 0 are:

  • x>7 x > -7
  • x<5 x < -5

Therefore, the correct answer is:

x>7 x > -7 or x<5 x < -5

3

Final Answer

x>7 x > -7 or x<5 x < -5

Key Points to Remember

Essential concepts to master this topic
  • Rule: Downward parabola is negative outside its roots
  • Technique: Find roots using quadratic formula: x=2±122=5,7 x = \frac{-2 \pm 12}{-2} = -5, -7
  • Check: Test intervals: f(8)=13<0 f(-8) = -13 < 0 and f(0)=35>0 f(0) = 35 > 0

Common Mistakes

Avoid these frequent errors
  • Getting interval directions wrong for downward parabolas
    Don't assume the parabola is negative between the roots = backwards answer! Since this parabola opens downward (negative leading coefficient), it's positive between roots and negative outside them. Always check the leading coefficient first to determine parabola direction.

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

How do I know if the parabola opens up or down?

+

Look at the leading coefficient (the number in front of x2 x^2 ). If it's positive, the parabola opens upward. If it's negative (like our -1), it opens downward.

Why is the answer 'x > -5 or x < -7' instead of between the roots?

+

Since our parabola opens downward, it's shaped like an upside-down U. This means it's positive between the roots (-7 and -5) and negative outside them.

What if I can't factor the quadratic?

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Use the quadratic formula! For ax2+bx+c=0 ax^2 + bx + c = 0 , the roots are x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} . This always works, even when factoring is difficult.

How do I test which intervals are negative?

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Pick any test value from each interval and substitute it into the original function. If the result is negative, that entire interval satisfies f(x)<0 f(x) < 0 .

What does 'or' mean in the final answer?

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The word 'or' means the function is negative in either interval. So x can be less than -7 OR greater than -5, but not both at the same time.

Do I include the roots in my answer for < 0?

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No! Since we want f(x)<0 f(x) < 0 (strictly less than), the roots where f(x)=0 f(x) = 0 are not included. Use parentheses or < and > symbols, not brackets or ≤ and ≥.

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