Question

j $3x^5 = 1$
3 solve for $x$:
a $(x - 3)^2 = 16$
d $(x - 7)^2 = 0$
★ $(x - \sqrt{2})^2 = 2$
4 solve for $x$:

Explanation:

Step1: Isolate squared term (part a)

$(x-3)^2 = 16$

Step2: Take square roots (part a)

$x-3 = \pm\sqrt{16} = \pm4$

Step3: Solve for x (part a)

$x = 3 + 4 = 7$ or $x = 3 - 4 = -1$

Step4: Isolate squared term (part d)

$(x-7)^2 = 0$

Step5: Take square root (part d)

$x-7 = \sqrt{0} = 0$

Step6: Solve for x (part d)

$x = 7$

Step7: Isolate squared term (starred)

$(x-\sqrt{2})^2 = 2$

Step8: Take square roots (starred)

$x-\sqrt{2} = \pm\sqrt{2}$

Step9: Solve for x (starred)

$x = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$ or $x = \sqrt{2} - \sqrt{2} = 0$

Answer:

a. $x=7$ or $x=-1$
d. $x=7$
Starred problem: $x=2\sqrt{2}$ or $x=0$