Question

solving quadratic inequalities
consider the quadratic inequality $x^2 - 1 > x + 5$.

what is the solution set to the given inequality?
a {x | x > 1 or x < -2}
b {x | x > 2 or x < -1}
c {x | x > 3 or x < -2}
d {x | x > 0 or x < 4}

what is the solution for $x^2 + 4 \leq 0$?
a $x > 0$
b $-2 < x < 2$
c $x > 2$
d no solution

Explanation:

First Question: Solve \( x^2 - 1 > x + 5 \)

Step1: Rearrange the inequality

Subtract \( x \) and \( 5 \) from both sides to get \( x^2 - x - 6 > 0 \).

Step2: Factor the quadratic

Factor \( x^2 - x - 6 \) as \( (x - 3)(x + 2) > 0 \).

Step3: Find critical points

Set \( (x - 3)(x + 2) = 0 \), so critical points are \( x = 3 \) and \( x = -2 \).

Step4: Test intervals

• For \( x < -2 \), say \( x = -3 \), \( (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 \), so this interval is part of the solution.
• For \( -2 < x < 3 \), say \( x = 0 \), \( (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 \), not part of the solution.
• For \( x > 3 \), say \( x = 4 \), \( (4 - 3)(4 + 2) = (1)(6) = 6 > 0 \), so this interval is part of the solution.

So the solution is \( \{x | x > 3 \text{ or } x < -2\} \), which is option C.

Second Question: Solve \( x^2 + 4 \leq 0 \)

Step1: Analyze \( x^2 \)

We know that for any real number \( x \), \( x^2 \geq 0 \). So \( x^2 + 4 \geq 4 \) for all real \( x \).

Step2: Determine solution

Since \( x^2 + 4 \) is always greater than \( 0 \) (in fact, at least \( 4 \)), there are no real numbers \( x \) that satisfy \( x^2 + 4 \leq 0 \). So the solution is no solution, which is option D.

Answer:

First Question: C. \(\{x\mid x > 3 \text{ or } x < -2\}\)
Second Question: D. no solution