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
$x = \boxed{\phantom{\text{1}}}$, $x = \boxed{\phantom{\text{1}}}$ submit answer

Explanation:

Step1: Isolate the squared term

$5(3x-3)^2 = -29 + 44$
$5(3x-3)^2 = 15$
$(3x-3)^2 = \frac{15}{5} = 3$

Step2: Take square root of both sides

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

Step3: Solve for x (positive root)

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

Step4: Solve for x (negative root)

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

*Note: Correction to initial miscalculation:

Step1: Isolate squared term

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

Step2: Square root both sides

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

Step3: Solve for positive case

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

Step4: Solve for negative case

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

Final corrected answer:

Step1: Isolate squared expression

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

Step2: Take square root of both sides

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

Step3: Solve for positive root

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

Step4: Solve for negative root

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

Answer:

$x = 2$, $x = 0$