Pre-Calculus: Find all real solutions of the equations

Question

Find all real solutions of the equations:

1.) (x²+3)²–4(x²+3)–5=0

2.)|x/2x–3|≥1

Answer 1

Hello, Louise,

1. (x²+3)²–4(x²+3)–5=0

x⁴₊6x² +9-4x² -12 -5 = 0

x⁴ + 2x² -8 = 0

x = - 2^1/2

x = 2^1/2

2.) |x/2x–3|≥1

I'm having difficulty interpreting this equation. It looks like the absolute value of ((x/2x)-3), as written. Is it actually x/(2x-3)? The first just reduces to [abs val (0.5-3)] = 2.5