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 Sub - Question (Solving \(x^{2}-1>x + 5\))

Step 1: Rearrange the inequality

First, we need to get all terms on one side of the inequality. Subtract \(x\) and \(5\) from both sides of \(x^{2}-1>x + 5\) to get a quadratic inequality in standard form.
\(x^{2}-x-6>0\)

Step 2: Factor the quadratic

We factor the quadratic expression \(x^{2}-x - 6\). We need two numbers that multiply to \(-6\) and add up to \(-1\). The numbers are \(-3\) and \(2\). So, \(x^{2}-x - 6=(x - 3)(x+2)\). So the inequality becomes \((x - 3)(x + 2)>0\)

Step 3: Find the critical points

The critical points are the values of \(x\) that make the expression equal to zero. Set \((x - 3)(x + 2)=0\). Then \(x-3 = 0\) gives \(x = 3\) and \(x+2=0\) gives \(x=-2\). These critical points divide the number line into three intervals: \((-\infty,-2)\), \((-2,3)\), and \((3,\infty)\)

Step 4: Test intervals

• For the interval \((-\infty,-2)\), let's pick \(x=-3\). Substitute into \((x - 3)(x + 2)\): \((-3-3)(-3 + 2)=(-6)(-1)=6>0\), so this interval is part of the solution.
• For the interval \((-2,3)\), let's pick \(x = 0\). Substitute into \((x - 3)(x + 2)\): \((0 - 3)(0+2)=(-3)(2)=-6<0\), so this interval is not part of the solution.
• For the interval \((3,\infty)\), let's pick \(x = 4\). Substitute into \((x - 3)(x + 2)\): \((4 - 3)(4+2)=(1)(6)=6>0\), so this interval is part of the solution.

So the solution to \(x^{2}-x - 6>0\) is \(x>3\) or \(x<-2\)

Second Sub - Question (Solving \(x^{2}+4\leq0\))

Step 1: Analyze the quadratic function

The function \(y=x^{2}+4\) is a quadratic function with a leading coefficient of \(1\) (positive) and a constant term of \(4\). The vertex form of a quadratic function is \(y=a(x - h)^{2}+k\), for \(y=x^{2}+4\), \(a = 1\), \(h = 0\), \(k = 4\). Since \(a>0\), the parabola opens upwards, and the minimum value of the function occurs at the vertex \((0,4)\).

Step 2: Determine the solution

We want to find when \(x^{2}+4\leq0\). But since the minimum value of \(x^{2}+4\) is \(4\) (when \(x = 0\)), and \(4>0\), there are no real values of \(x\) for which \(x^{2}+4\leq0\). So the solution is no solution.

Answer:

s:
For the first sub - question (solving \(x^{2}-1>x + 5\)):