Question

compound and absolute value inequalities
score: 3/10 penalty: 1 off
question
solve the following inequality algebraically.
$3x^2 - 44x + 151 \leq -2x + 7$
answer: interval: (use of x

Explanation:

Step1: Rearrange to standard quadratic form

Move all terms to left side:
$3x^2 - 44x + 151 + 2x - 7 \leq 0$
Simplify: $3x^2 - 42x + 144 \leq 0$

Step2: Simplify the quadratic inequality

Divide all terms by 3:
$\frac{3x^2 - 42x + 144}{3} \leq \frac{0}{3}$
Result: $x^2 - 14x + 48 \leq 0$

Step3: Factor the quadratic

Find two numbers that multiply to 48 and add to -14:
$(x - 6)(x - 8) \leq 0$

Step4: Identify critical points

Set factors equal to 0:
$x - 6 = 0 \implies x=6$; $x - 8 = 0 \implies x=8$

Step5: Test intervals for sign

Test $x<6$ (e.g., $x=5$): $(5-6)(5-8)=(-1)(-3)=3>0$ (not valid)
Test $6Test $x>8$ (e.g., $x=9$): $(9-6)(9-8)=(3)(1)=3>0$ (not valid)
Check critical points: $(6-6)(6-8)=0$, $(8-6)(8-8)=0$ (valid, as inequality is $\leq0$)

Answer:

$6 \leq x \leq 8$