Solve for x: (2x - 6)(x + 5) = 0; 7x^2 - 11x + 3 = 0 (correct to TWO decimal places); x^2 ≥ 5x; 3√(x + 12) - x = 8.

Steps to Solve

1. Solve the first equation: $(2x - 6)(x + 5) = 0$

This equation is set to zero, so we can use the Zero Product Property. Set each factor to zero:

$$ 2x - 6 = 0 $$ and $$ x + 5 = 0 $$

Solving these gives:

$$ 2x = 6 \implies x = 3 $$

and

$$ x = -5 $$

The solutions are $x = 3$ and $x = -5$.

2. Solve the second equation: $7x^2 - 11x + 3 = 0$

To solve this quadratic equation, we can use the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where $a = 7$, $b = -11$, and $c = 3$.

Calculating the discriminant:

$$ b^2 - 4ac = (-11)^2 - 4(7)(3) = 121 - 84 = 37 $$

Now substituting into the quadratic formula:

$$ x = \frac{11 \pm \sqrt{37}}{14} $$

Approximating the solutions to two decimal places gives:

$$ x \approx \frac{11 + 6.08}{14} \approx 1.23 $$

$$ x \approx \frac{11 - 6.08}{14} \approx 0.35 $$

The solutions are approximately $x \approx 1.23$ and $x \approx 0.35$.

3. Solve the inequality: $x^2 \geq 5x$

Rearranging gives:

$$ x^2 - 5x \geq 0 $$

Factoring out:

$$ x(x - 5) \geq 0 $$

The critical points are $x = 0$ and $x = 5$. Analyzing intervals, we find:

• For $(-\infty, 0)$: Positive
• For $(0, 5)$: Negative
• For $(5, \infty)$: Positive

Thus, the solution is:

$$ x \in (-\infty, 0] \cup [5, \infty) $$

4. Solve the equation: $3\sqrt{x + 12} - x = 8$

First, isolate the square root:

$$ 3\sqrt{x + 12} = x + 8 $$

Now divide by 3:

$$ \sqrt{x + 12} = \frac{x + 8}{3} $$

Squaring both sides yields:

$$ x + 12 = \left(\frac{x + 8}{3}\right)^2 $$

Expanding gives:

$$ x + 12 = \frac{(x + 8)^2}{9} $$

Multiplying through by 9:

$$ 9(x + 12) = (x + 8)^2 $$

This simplifies to:

$$ 9x + 108 = x^2 + 16x + 64 $$

Rearranging gives:

$$ x^2 + 7x - 44 = 0 $$

Using the quadratic formula again:

$$ x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-44)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 176}}{2} = \frac{-7 \pm \sqrt{225}}{2} $$

Calculating gives:

$$ x = \frac{-7 \pm 15}{2} $$

This results in:

$$ x = 4 \quad \text{or} \quad x = -11 $$

Now, since $x + 12 \geq 0$ in the square root, we reject $x = -11$.

The solution is $x = 4$.