Finding X: Solve y = x² + 5x + 4 for Negative Output

Quadratic Inequalities with Sign Analysis

Look at the following function:

y=x2+5x+4 y=x^2+5x+4

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+5x+4 y=x^2+5x+4

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 the function f(x)=x2+5x+4 f(x) = x^2 + 5x + 4 is less than zero, we will first factor the quadratic expression.

Step 1: Factor the quadratic function.

  • The quadratic x2+5x+4 x^2 + 5x + 4 can be factored into (x+4)(x+1) (x + 4)(x + 1) .

Step 2: Find the roots of the quadratic equation.

  • Set each factor equal to zero: x+4=0 x + 4 = 0 and x+1=0 x + 1 = 0 .
  • The solutions are x=4 x = -4 and x=1 x = -1 .

Step 3: Determine the sign of the quadratic in the intervals defined by these roots.

  • Consider the intervals (,4) (-\infty, -4) , (4,1) (-4, -1) , and (1,) (-1, \infty) .
  • Pick test points from each interval: for (,4) (-\infty, -4) , choose x=5 x = -5 ; for (4,1) (-4, -1) , choose x=2 x = -2 ; for (1,) (-1, \infty) , choose x=0 x = 0 .
  • Evaluate the sign of (x+4)(x+1) (x+4)(x+1) at each test point:
    • At x=5 x = -5 , (x+4)(x+1)=(1)(4)=4 (x+4)(x+1) = (-1)(-4) = 4 (positive).
    • At x=2 x = -2 , (x+4)(x+1)=(2)(1)=2 (x+4)(x+1) = (2)(-1) = -2 (negative).
    • At x=0 x = 0 , (x+4)(x+1)=(4)(1)=4 (x+4)(x+1) = (4)(1) = 4 (positive).
  • Therefore, the function is negative in the interval (4,1) (-4, -1) .

Consequently, the solution is 4<x<1 -4 < x < -1 .

The correct choice from the options given is choice 4.

3

Final Answer

4<x<1 -4 < x < -1

Key Points to Remember

Essential concepts to master this topic
  • Factoring: Factor quadratic expression to find where it equals zero
  • Technique: Test signs in intervals: x2+5x+4=(x+4)(x+1) x^2 + 5x + 4 = (x+4)(x+1)
  • Check: Substitute test point x=2 x = -2 : (2)2+5(2)+4=2<0 (-2)^2 + 5(-2) + 4 = -2 < 0

Common Mistakes

Avoid these frequent errors
  • Solving f(x) = 0 instead of f(x) < 0
    Don't just find the roots x = -4 and x = -1 and stop there = incomplete answer! Finding roots only tells you where the parabola crosses the x-axis, not where it's negative. Always test intervals between roots to determine sign changes.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the \( x \)-axis.

The parabola's vertex is marked A.

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

AAAX

FAQ

Everything you need to know about this question

Why do I need to factor the quadratic first?

+

Factoring reveals the roots of the quadratic, which are the boundary points where the function changes from positive to negative (or vice versa). These roots divide the number line into intervals you can test.

How do I know which intervals to test?

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The roots divide the number line into sections. For x=4 x = -4 and x=1 x = -1 , test one point in each interval: before -4, between -4 and -1, and after -1.

What if I can't factor the quadratic easily?

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Use the quadratic formula to find the roots first: x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} . Then proceed with interval testing just like with factored form.

Why is the answer between the roots and not outside them?

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Since the coefficient of x2 x^2 is positive (it's 1), this parabola opens upward. This means it's negative between its roots and positive outside them.

How can I double-check my interval answer?

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Pick any number in your answer interval and substitute it into the original function. For example, x=2.5 x = -2.5 : (2.5)2+5(2.5)+4=2.25<0 (-2.5)^2 + 5(-2.5) + 4 = -2.25 < 0

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