Question

Find the values of \(\mathrm{y}\) which satisfy the quadratic inequalities below \(\mathrm{y}^{2}+5 \mathrm{y}+4 \leq 0\) and \(\mathrm{y}^{2}-2 \mathrm{y}-\) \(15 \geq 0\) (1) \(-1 \leq \mathrm{y} \leq 5\) (2) \(-4 \leq \mathrm{y} \leq-3\) (3) Either (1) or (2) (4) Neither (1) nor (2)

Short answer

(1) \(y \leq -5\) (2) \(-4 \leq y \leq -3\) (3) \(-3 \leq y \leq -1\) (4) \(y \geq 5\) Answer: Option (2) \(-4 \leq y \leq -3\).

Step-by-step solution

Solve the first inequality

To solve the first inequality \(y^2+5y+4 \leq 0\), we'll start by finding its critical points. By factoring, we can rewrite the equation as \((y+4)(y+1) = 0\). So, the critical points are \(y=-4\) and \(y=-1\). Check the intervals \((-\infty, -4], [-4, -1]\) and \([-1, \infty)\) to determine which of these intervals satisfy the inequality, by choosing any value from those intervals and evaluate: \((-\infty, -4]:\) Choose \(y=-5,\) we get \((-5+4)(-5+1) = (-1)(-4)\) which is positive, therefore it doesn't satisfy the inequality. \([-4,-1]:\) Choose \(y=-2,\) we get \((-2+4)(-2+1) = (2)(-1)\) which is negative, therefore it satisfies the inequality. \([-1, \infty):\) Choose \(y=0,\) we get \((0+4)(0+1) = (4)(1)\) which is positive, therefore it doesn't satisfy the inequality. Hence, the first inequality holds true when \(-4 \leq y \leq -1\).

Solve the second inequality

To solve the second inequality \(y^2-2y-15\geq0\), we'll start by finding its critical points. By factoring, we can rewrite the equation as \((y-5)(y+3)=0\). So, the critical points are \(y=5\) and \(y=-3\). Check the intervals \((-\infty, -3], [-3, 5]\) and \([5, \infty)\) to determine which of these intervals satisfy the inequality, by choosing any value from those intervals and evaluate: \((-\infty, -3]:\) Choose \(y=-4,\) we get \((-4-5)(-4+3) = (-9)(-1)\) which is positive, therefore it satisfies the inequality. \([-3, 5]:\) Choose \(y=0,\) we get \((0-5)(0+3) = (-5)(3)\) which is negative, therefore it doesn't satisfy the inequality. \([5, \infty):\) Choose \(y=6,\) we get \((6-5)(6+3) = (1)(9)\) which is positive, therefore it satisfies the inequality. Hence, the second inequality holds true when \(y \leq -3\) or \(y \geq 5\).

Determine the common solution

Now we want to find the range of y-values that satisfy both inequalities. The first inequality is satisfied by \(-4 \leq y \leq -1\), while the second inequality is satisfied by \(y \leq -3\) or \(y \geq 5\). Comparing the ranges, we can see that only the range \(-4 \leq y \leq -3\) satisfies both inequalities. Looking at the given options, we can see this range is found in option (2). So, the correct answer is \((2) \: -4 \leq y \leq -3\).

Key concepts

Critical Points

In solving quadratic inequalities, identifying critical points is a crucial step. These points are the solutions of the corresponding quadratic equation where the inequality becomes an equality. For any quadratic inequality of the form \(ax^2 + bx + c \geq 0\) or \(ax^2 + bx + c \leq 0\), the critical points are found by setting the equation \(ax^2 + bx + c = 0\) and solving for \y\.

In our examples, the critical points for the inequalities \(y^2 + 5y + 4 \leq 0\) and \(y^2 - 2y - 15 \geq 0\) are calculated by finding \(y = -4, -1\) and \(y = -3, 5\) respectively after factorization. These critical points partition the number line into intervals that we need to test for the inequality conditions. Critical points are essential as they narrow down the intervals where inequalities might hold true.

Inequality Range

The inequality range determines the values that satisfy a specific quadratic inequality on a number line. By evaluating intervals determined by the critical points, you can establish where the inequality is true.

For these quadratic inequalities, each partition formed by the critical points is tested to see if they satisfy the inequality. For instance, \(-4 \leq y \leq -1\) satisfies the first inequality \(y^2 + 5y + 4 \leq 0\). Meanwhile, the second inequality \(y^2 - 2y - 15 \geq 0\) is satisfied in the intervals \(y \leq -3\) or \(y \geq 5\). The calculations at each interval will show whether expressions evaluate to true or false for that section. Hence, the final answer for both inequalities lies within the overlap of their satisfying intervals.

Factorization

Factorization is a technique used to simplify polynomials by expressing them as a product of their factors. It is handy in solving quadratic equations like \(y^2 + 5y + 4 = 0\) and \(y^2 - 2y - 15 = 0\). Factorizing these equations simplifies finding critical points and solving inequalities.

For \(y^2 + 5y + 4 \leq 0\), factorization gives \((y + 4)(y + 1) = 0\), yielding critical points at \(-4\) and \(-1\). Similarly, \(y^2 - 2y - 15 \geq 0\) can be rewritten as \((y - 5)(y + 3) = 0\), leading to critical points at \(5\) and \(-3\).

Factorization simplifies checking intervals for inequalities efficiently and allows us to quickly determine where the expressions evaluate to zero or change signs.

Interval Checking

Performing interval checking is crucial to determine which portions of a quadratic inequality hold true. After finding the critical points and the factorization, you begin checking each interval delimited by these critical points. You pick test points in each interval, substitute into the factored form of the inequality, and determine if the expression results in a true statement (either non-positive or non-negative depending on the inequality).

For the inequality \(y^2 + 5y + 4 \leq 0\), intervals are checked as follows:

• Choose a value less than \-4\, like \y = -5\. It results positive, meaning the inequality does not hold.
• Pick a value between \(-4, -1\), say \y = -2\. It results in a negative product, satisfying the condition.
• A value greater than \-1,\ such as \y = 0\ results positive, not satisfying the inequality.

A similar process occurs for the second inequality \(y^2 - 2y - 15 \geq 0\), allowing confident determination of valid solution sets. This technique is fundamental for efficiently solving and understanding quadratic inequalities.