Find the complete solution set

Didn't understand professors lecture at all. -_- Could someone explain to me in detail these 2 problems...

(x^4)+4=5(x^2)

log2(x) + log2(3x + 10) = 3

1) x^4 - 5x^2 + 4 = 0 (minor but critical rearrangement of polynomial).

Let u = x^2 (common trick)

Now we have: u^2 - 5u + 4 = 0.............factor

(u - 4) * (u - 1) = 0

u = 1 and u = 4

Since u = x^2, x = +/- sqrt(u)

{+1, -1, +2, -2} <---------------------- Solution set for #1

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2) log2(x) + log2(3x + 10) = 3

There is a basic log(regardless of base) that states (log(a)) + (log(b)) = log(a*b)

Apply that to the LHS

log2(3x^2 + 10x) = 3

Raise 2 to the power of both sides of the equation.

(2^(log2(3x^2 + 10x))) = 2^3 = 8

The power and log cancel each other so we now have:

3x^2 + 10x = 8

3x^2 + 10x - 8 = 0

(3x + 12) * (x - (2/3)) = 0 (or use the quadratic formula)

3x + 12 = 0 implies that x = -4

(x - (2/3)) = 0 implies that x = (2/3)

{ -4, (2/3) } <----------- Solution set for #2

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