Question

8.6 solve radical equations (homework)
score: 7/16 answered: 7/16
question 8
solve: $sqrt{q + 36}-q + 6 = 0$
$q=$
question help: video written example
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Explanation:

Step1: Isolate the radical term

$\sqrt{q + 36}=q - 6$

Step2: Square both sides

$(\sqrt{q + 36})^2=(q - 6)^2$
$q + 36=q^{2}-12q + 36$

Step3: Rearrange to form a quadratic equation

$q^{2}-12q + 36-(q + 36)=0$
$q^{2}-12q + 36 - q-36=0$
$q^{2}-13q=0$

Step4: Factor the quadratic equation

$q(q - 13)=0$

Step5: Solve for q

$q = 0$ or $q-13=0$, so $q = 0$ or $q=13$

Step6: Check for extraneous solutions

When $q = 0$: $\sqrt{0 + 36}-0 + 6=\sqrt{36}+6=6 + 6=12
eq0$, so $q = 0$ is extraneous.
When $q = 13$: $\sqrt{13+36}-13 + 6=\sqrt{49}-7=7 - 7=0$

Answer:

$13$