Question

question #9 a student is trying to solve the quadratic inequality -2x² - 7x < -4. the students work is shown. did the student make a mistake? if so, in which step was the mistake made? what was the mistake? step 1: -2x² - 7x + 4 < 0 step 2: 2x² + 7x - 4 > 0 step 3: (2x - 1)(x + 4) > 0 step 4: 2x - 1 = 0, x = 1/2; x + 4 = 0, x = -4 step 5: when x = -5: -2x² - 7x < -4, -2(-5)² - 7(-5) < -4, -50 + 35 < -4, true; x = 0: -2x² - 7x < -4, -2(0)² - 7(0) < -4, 0 + 0 < -4, false; x = 1: -2x² - 7x < -4, -2(1)² - 7(1) < -4, -2 - 7 < -4, true step 6: solution set (-∞, -4 ∪ 1/2, ∞) or x ≤ -4 x ≥ 1/2 step 2: the student has an incorrect inequality symbol. when dividing an inequality by a negative number you do not have to flip the inequality symbol. step 5: when evaluating test points, the student made arithmetic errors making the inequalities incorrect. step 6: the student wrote the solution set incorrectly. the solution set is (-∞, -4) ∪ (1/2, ∞) which is the same as x < -4 or x > 1/2 no errors were made. the work is correct

Explanation:

Step1: Analyze Step 2

When going from \(-2x^{2}-7x + 4<0\) to \(2x^{2}+7x - 4>0\), we multiply the entire inequality by - 1. According to the rules of inequalities, when we multiply or divide an inequality by a negative number, the inequality sign flips. This step is correct.

Step2: Analyze Step 5

For \(x=-5\):
\[

$$\begin{align*} -2x^{2}-7x&=-2\times(-5)^{2}-7\times(-5)\\ &=-2\times25 + 35\\ &=-50+35\\ &=-15<-4 \end{align*}$$

\]
The statement \(-50 + 35<-4\) is correct.
For \(x = 0\):
\[

$$\begin{align*} -2x^{2}-7x&=-2\times0^{2}-7\times0\\ &=0 ot<-4 \end{align*}$$

\]
The statement \(0 + 0<-4\) is correct in terms of the inequality result.
For \(x = 1\):
\[

$$\begin{align*} -2x^{2}-7x&=-2\times1^{2}-7\times1\\ &=-2-7\\ &=-9<-4 \end{align*}$$

\]
The statement \(-2 - 7<-4\) is correct. So, there are no arithmetic - errors in step 5.

Step3: Analyze Step 6

The roots of the quadratic equation \(2x^{2}+7x - 4=0\) are \(x=\frac{1}{2}\) and \(x=-4\). The quadratic function \(y = 2x^{2}+7x - 4\) is a parabola opening upwards (since the coefficient of \(x^{2}\) is positive, \(a = 2>0\)). The inequality \(2x^{2}+7x - 4>0\) is satisfied when \(x<-4\) or \(x>\frac{1}{2}\), which is written in interval notation as \((-\infty,-4)\cup(\frac{1}{2},\infty)\). The student wrote the solution set with incorrect brackets. The solution set should be open - ended at \(-4\) and \(\frac{1}{2}\) because the original inequality \(-2x^{2}-7x<-4\) (or \(2x^{2}+7x - 4>0\)) does not include the equal - to case.

Answer:

Step 6; The student wrote the solution set incorrectly. The solution set is \((-\infty,-4)\cup(\frac{1}{2},\infty)\) which is the same as \(x < - 4\) or \(x>\frac{1}{2}\)