Question

fva 25 - 26: algebra ii q3
cubic and cube root functions and equations
what is the solution to the equation $sqrt{n}=sqrt{3n - 3}+1$
$n=-4, n=-1$
$n=1, n=4$
$n=1$
$n=-1$

Explanation:

Step1: Isolate one radical

$$\sqrt{n} - 1 = \sqrt{3n - 3}$$

Step2: Square both sides

$$(\sqrt{n} - 1)^2 = (\sqrt{3n - 3})^2$$
$$n - 2\sqrt{n} + 1 = 3n - 3$$

Step3: Isolate remaining radical

$$-2\sqrt{n} = 2n - 4$$
$$\sqrt{n} = -n + 2$$

Step4: Square both sides again

$$(\sqrt{n})^2 = (-n + 2)^2$$
$$n = n^2 - 4n + 4$$

Step5: Rearrange to quadratic equation

$$n^2 - 5n + 4 = 0$$

Step6: Factor quadratic

$$(n - 1)(n - 4) = 0$$
$$n=1 \text{ or } n=4$$

Step7: Verify solutions

For $n=1$: $\sqrt{1} = \sqrt{3(1)-3}+1 \implies 1=0+1$, valid.
For $n=4$: $\sqrt{4} = \sqrt{3(4)-3}+1 \implies 2=3+1$, invalid.

Answer:

n=1