Question

solve the following quadratic equation for all values of $x$ in simplest form.
$5(3x - 3)^2 - 44 = -29$
answer attempt 2 out of 2
additional solution remove solution
$x = \boldsymbol{\boxed{}}$, $x = \boldsymbol{\boxed{}}$ submit answer

Explanation:

Step1: Isolate the squared term

Add 44 to both sides:
$$5(3x-3)^2 = -29 + 44$$
$$5(3x-3)^2 = 15$$

Step2: Simplify the equation

Divide both sides by 5:
$$(3x-3)^2 = 3$$

Step3: Take square roots of both sides

$$3x-3 = \pm\sqrt{3}$$

Step4: Solve for x (positive root)

Add 3, then divide by 3:
$$3x = 3 + \sqrt{3}$$
$$x = \frac{3+\sqrt{3}}{3} = 1 + \frac{\sqrt{3}}{3}$$
Wait, correction: Original calculation error. Let's redo Step1-4 correctly:

Step1: Isolate squared term

$$5(3x-3)^2 = -29 + 44$$
$$5(3x-3)^2 = 15$$

Step2: Divide by 5

$$(3x-3)^2 = 3$$

Step3: Square root both sides

$$3x-3 = \pm\sqrt{3}$$

Step4: Solve for x (case 1: +√3)

$$3x = 3 + \sqrt{3}$$
$$x = \frac{3+\sqrt{3}}{3} = 1 + \frac{\sqrt{3}}{3}$$

Step5: Solve for x (case 2: -√3)

$$3x = 3 - \sqrt{3}$$
$$x = \frac{3-\sqrt{3}}{3} = 1 - \frac{\sqrt{3}}{3}$$

Wait, rechecking the original equation: $5(3x-3)^2 -44 = -29$
$5(3x-3)^2 = 15$
$(3x-3)^2 = 3$
$3x-3 = \sqrt{3}$ or $3x-3 = -\sqrt{3}$
$3x = 3+\sqrt{3}$ → $x = 1 + \frac{\sqrt{3}}{3}$
$3x = 3-\sqrt{3}$ → $x = 1 - \frac{\sqrt{3}}{3}$

Final corrected answer:

Step1: Isolate squared expression

Add 44 to both sides:
$$5(3x-3)^2 = 15$$

Step2: Simplify to get squared term

Divide by 5:
$$(3x-3)^2 = 3$$

Step3: Take square roots

$$3x-3 = \pm\sqrt{3}$$

Step4: Solve for positive root

Add 3, divide by 3:
$$x = \frac{3+\sqrt{3}}{3} = 1+\frac{\sqrt{3}}{3}$$

Step5: Solve for negative root

Add 3, divide by 3:
$$x = \frac{3-\sqrt{3}}{3} = 1-\frac{\sqrt{3}}{3}$$

Answer:

$x=2$, $x=0$