Question

question #1 a student is trying to solve the quadratic inequality 3x² + 5x < 2. 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: 3x² + 5x - 2 < 0 step 2: (3x - 1)(x + 2) < 0 step 3: 3x - 1 = 0 x + 2 = 0 x = 1/3 x = -2 step 4: x = -3: 3x² + 5x < 2 3(-3)² + 5(-3) < 2 27 - 15 < 2 false x = 0: 3x² + 5x < 2 3(0)² + 5(0) < 2 0 + 0 < 2 true when x = 1: 3x² + 5x < 2 3(1)² + 5(1) < 2 3 + 5 < 2 false step 5: solution: (-2 1/3) or -2 < x < 1/3 step 5; the student wrote the solution set incorrectly. the solution set is (1/3, -2) or 1/3 < x < -2 step 2; the student has the wrong factors. the factors should be (2x - 1)(x + 3). no errors were made. the work is correct. step 4; when evaluating test points, the student made arithmetic errors making the inequalities incorrect.

Explanation:

Step1: Analyze Step 1

Subtract 2 from both sides of $3x^{2}+5x < 2$ to get $3x^{2}+5x - 2<0$. This step is correct.

Step2: Analyze Step 2

Factor $3x^{2}+5x - 2$ as $(3x - 1)(x + 2)<0$. Using FOIL: $(3x-1)(x + 2)=3x\times x+3x\times2-1\times x - 1\times2=3x^{2}+6x - x-2=3x^{2}+5x - 2$. This step is correct.

Step3: Analyze Step 3

Set each factor equal to zero: $3x - 1 = 0$ gives $x=\frac{1}{3}$ and $x + 2=0$ gives $x=-2$. These are the critical - points and this step is correct.

Step4: Analyze Step 4

For $x=-3$: $3(-3)^{2}+5(-3)=3\times9-15 = 27 - 15=12$, and $12
ot<2$.
For $x = 0$: $3(0)^{2}+5(0)=0$ and $0<2$.
For $x = 1$: $3(1)^{2}+5(1)=3 + 5=8$, and $8
ot<2$. The arithmetic in step 4 for evaluating test - points is correct.

Step5: Analyze Step 5

The inequality $(3x - 1)(x + 2)<0$ is negative when $-2

Answer:

No errors were made. The work is correct.