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Chapter 0: Problem 38

Solve each polynomial inequality and express the solution set in interval notation. $$8 s+12 \leq-s^{2}$$

Short Answer

Expert verified
The solution set for the inequality is \([-6, -2]\).

Step by step solution

01

Rewrite the Inequality

To solve the inequality, we first rewrite it in a standard form where one side is zero. Start with the given inequality: \(8s + 12 \leq -s^2\). Add \(s^2\) to both sides to get: \(s^2 + 8s + 12 \leq 0\).
02

Find the Values to Test

Factor the quadratic expression \(s^2 + 8s + 12\). After testing possible factors, factor it as \((s + 2)(s + 6) = 0\). This gives roots \(s = -2\) and \(s = -6\). These roots divide the number line into intervals to test in the inequality.
03

Test Intervals

The roots \(s = -2\) and \(s = -6\) create three intervals: \(( -\infty, -6 ), \), \(( -6, -2 ), \), and \(( -2, \infty )\). Choose test points from each interval: \(s = -7\) (from \( -\infty, -6 \)), \(s = -4\) (from \( -6, -2 \)), and \(s = 0\) (from \( -2, \infty \)).
04

Determine Where the Inequality Holds

Substitute the test points into \((s + 2)(s + 6)\leq 0\):- For \(s = -7\), \(( -7 + 2)( -7 + 6) = 5(-1) = -5\), satisfies the inequality.- For \(s = -4\), \((-4 + 2)(-4 + 6) = -2(2) = -4\), satisfies the inequality.- For \(s = 0\), \((0 + 2)(0 + 6) = 2 \times 6 = 12\), does not satisfy the inequality.
05

Confirm the Solution

The inequality holds for \(-\infty < s \leq -2\) and \(s = -6\). Therefore, the solution to the inequality \(s^2 + 8s + 12 \leq 0\) in interval notation is \([-6, -2]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Quadratic Inequalities
Quadratic inequalities involve expressions like \( ax^2 + bx + c \leq 0 \), where \( a \), \( b \), and \( c \) are constants. Solving these inequalities means finding the range of values for \( x \) that make the inequality true.
Here's a simple approach to tackle them:
  • First, rewrite the inequality so one side equals zero. This helps focus on the roots and the changes in sign across intervals.
  • Second, factor the quadratic expression if possible. This helps identify the roots, which are critical points where the sign of the expression might change.
  • Third, use the roots to divide the number line into intervals. You'll test each interval to see where the inequality holds.
Quadratic inequalities can seem daunting, but by systematically dividing the number line and testing points, you can determine where the inequality holds true.
Expressing Solutions in Interval Notation
Interval notation is a concise way to express the set of solutions for an inequality. It describes the range of numbers that satisfy the inequality.
When using interval notation:
  • Use brackets \([\;]\) when the endpoint is included in the solution set. If at \( x = a \), the condition is equal (\( \leq \) or \( \geq \)), then \([\; a \;]\) is used.
  • Use parentheses \((\; )\) when the endpoint is not included. For example, \( (a, b) \) means the values between \( a \) and \( b \) are included, but not \( a \) or \( b \).
  • Symbols \( -\infty \) and \( \infty \) are always enclosed in parentheses, as they represent unbounded intervals.
An example would be the interval \([-6, -2]\) which indicates that \( s \) includes all values from \(-6\) to \(-2\) inclusive.
Factoring Quadratics
To solve a quadratic inequality, factoring the quadratic expression can be a useful technique. This involves rewriting the quadratic as a product of its factors.
Here鈥檚 how:
  • Identify or determine two numbers that multiply to the constant term \( c \) and add to the linear coefficient \( b \).
  • Use these numbers to write the quadratic expression as a product of two binomials.
In the given example, the quadratic \( s^2 + 8s + 12 \) is factored as \((s + 2)(s + 6)\). This factorization reveals the roots \( s = -2 \) and \( s = -6 \), which help divide the number line into intervals to test the inequality.
Factoring is a foundational skill for solving quadratic inequalities and can simplify the process significantly.

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Most popular questions from this chapter

Solve the equation \(3(x+1)+2=x-3(x-1)\) Solution: $$ \begin{aligned} 3 x+3+2 &=x-3 x-3 \\ 3 x+5 &=-2 x-3 \\ 5 x &=-8 \\ x &=-\frac{8}{5} \end{aligned} $$ This is incorrect. What mistake was made?

Explain the mistake that is made. Find the slope of the line that passes through the points (-2,3) and (4,1) Solution: Write the slope formula. \(\quad m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) Substitute (-2,3) and \((4,1), \quad m=\frac{1-3}{-2-4}=\frac{-2}{-6}=\frac{1}{3}\) This is incorrect. What mistake was made?

Show that two lines with equal slopes and different \(y\) -intercepts have no point in common. Hint: Let \(y_{1}=m x+b_{1}\) and \(y_{2}=m x+b_{2}\) with \(b_{1} \neq b_{2} .\) What equation must be true for there to be a point of intersection? Show that this leads to a contradiction.

Solve the equation using factoring by grouping: \(x^{3}+2 x^{2}-x-2=0\)

Solve for the indicated variable in terms of other variables. Solve \(s=\frac{1}{2} g t^{2}\) for \(t\)
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