I think it is false sorry if i am wrong
The quadratic equation has two solutions if b^2 - 4ac > 0
Given the equation ax^4 + bx^2 + c=0
Substitute into the formula to have:
The equation becomes aP^2 + bP + c = 0
For us to have a unique solution, the discriminant b^2 - 4ac must be greater than zero. Hence the quadratic equation has two solutions if b^2 - 4ac > 0
learn more on discriminant here; brainly.com/question/1537997
Condense the logarithm on the right side:




So,

and
, which makes (4) the correct choice.
<h3>
Answer: x+4</h3>
This is because the given expression factors to (x+4)(x+4), which condenses to (x+4)^2.
To factor, think of two numbers that A) multiply to 16, and B) add to 8. Those values would be 4 and 4
4+4 = 8
4*4 = 16
So that's how we end up with (x+4)(x+4). You can use the FOIL rule to expand that out and get x^2+8x+16 again to help verify you have the correct factorization.