Question

3.5- 3.6 practice cw/hw
part i: solving inequalities in one variable practice

1. $0 < 100 - 25x^2$
2. $-3x^2 \leq 5 - 8x$
3. $-2x + 1 \leq -x^2$
4. $x^2 + 3x + 7 \geq 0$

Explanation:

Problem 1: \( 0 < 100 - 25x^2 \)

Step 1: Rearrange the inequality

Rewrite the inequality as \( 25x^2 < 100 \).

Step 2: Divide both sides by 25

We get \( x^2 < 4 \).

Step 3: Solve the quadratic inequality

Recall that \( x^2 < a \) (where \( a>0 \)) implies \( - \sqrt{a} < x < \sqrt{a} \). Here, \( a = 4 \), so \( - 2 < x < 2 \).

Step 1: Rearrange the inequality

Bring all terms to one side: \( -3x^2 + 8x - 5 \leq 0 \). Multiply both sides by - 1 (and reverse the inequality sign) to get \( 3x^2 - 8x + 5 \geq 0 \).

Step 2: Factor the quadratic

Factor \( 3x^2 - 8x + 5 \). We find two numbers that multiply to \( 3\times5 = 15 \) and add to - 8. The numbers are - 3 and - 5. So, \( 3x^2 - 3x - 5x + 5 = 3x(x - 1)-5(x - 1)=(3x - 5)(x - 1) \geq 0 \).

Step 3: Find critical points

Set each factor equal to zero: \( 3x - 5 = 0 \) gives \( x=\frac{5}{3} \), and \( x - 1 = 0 \) gives \( x = 1 \).

Step 4: Test intervals

The critical points divide the number line into three intervals: \( (-\infty, 1) \), \( (1,\frac{5}{3}) \), and \( (\frac{5}{3},\infty) \).

• For \( x < 1 \) (e.g., \( x = 0 \)): \( (3(0)-5)(0 - 1)=(-5)(-1)=5\geq0 \), so this interval is part of the solution.
• For \( 1 < x < \frac{5}{3} \) (e.g., \( x = 2 \)): \( (3(2)-5)(2 - 1)=(6 - 5)(1)=1\geq0 \)? Wait, no, \( x = 1.5 \) (which is between 1 and \( \frac{5}{3}\approx1.666 \)): \( (3(1.5)-5)(1.5 - 1)=(4.5 - 5)(0.5)=(-0.5)(0.5)=-0.25<0 \), so this interval is not part of the solution.
• For \( x > \frac{5}{3} \) (e.g., \( x = 2 \)): \( (3(2)-5)(2 - 1)=(6 - 5)(1)=1\geq0 \), so this interval is part of the solution. Also, the critical points satisfy the inequality (since it's \( \geq0 \)), so the solution is \( x\leq1 \) or \( x\geq\frac{5}{3} \).

Step 1: Rearrange the inequality

Bring all terms to one side: \( x^2 - 2x + 1 \leq 0 \).

Step 2: Factor the quadratic

Notice that \( x^2 - 2x + 1=(x - 1)^2 \). So, we have \( (x - 1)^2 \leq 0 \).

Step 3: Analyze the square

A square of a real number is always non - negative. The only time \( (x - 1)^2 = 0 \) is when \( x = 1 \). And since a square is never negative, the only solution is \( x = 1 \).

Answer:

\( -2 < x < 2 \)

Problem 2: \( -3x^2 \leq 5 - 8x \)