Question

solve for x.
$5x^2 - 30x + 25 > 0$
write a compound inequality like $1 < x < 3$ or like $x < 1$ or $x > 3$.

Explanation:

Step1: Factor out the common factor

First, factor out the greatest common factor from the quadratic expression. The GCF of \(5x^2\), \(-30x\), and \(25\) is \(5\). So we have:

$$ 5(x^2 - 6x + 5)>0 $$

Step2: Factor the quadratic inside the parentheses

Factor the quadratic \(x^2 - 6x + 5\). We need two numbers that multiply to \(5\) and add up to \(-6\). Those numbers are \(-1\) and \(-5\). So:

$$ x^2 - 6x + 5=(x - 1)(x - 5) $$

Now our inequality becomes:

$$ 5(x - 1)(x - 5)>0 $$

Since \(5>0\), we can divide both sides by \(5\) without changing the direction of the inequality:

$$ (x - 1)(x - 5)>0 $$

Step3: Find the critical points

The critical points are the values of \(x\) that make the expression equal to zero. Set \((x - 1)(x - 5)=0\). Then \(x - 1 = 0\) gives \(x = 1\) and \(x - 5 = 0\) gives \(x = 5\). These critical points divide the number line into three intervals: \((-\infty, 1)\), \((1, 5)\), and \((5, \infty)\).

Step4: Test intervals

• For the interval \((-\infty, 1)\), let's pick \(x = 0\). Substitute into \((x - 1)(x - 5)\): \((0 - 1)(0 - 5)=(-1)(-5)=5>0\). So the inequality is true in this interval.
• For the interval \((1, 5)\), let's pick \(x = 3\). Substitute into \((x - 1)(x - 5)\): \((3 - 1)(3 - 5)=(2)(-2)=-4<0\). So the inequality is false in this interval.
• For the interval \((5, \infty)\), let's pick \(x = 6\). Substitute into \((x - 1)(x - 5)\): \((6 - 1)(6 - 5)=(5)(1)=5>0\). So the inequality is true in this interval.

Answer:

\(x < 1\) or \(x > 5\)