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

Hi Louise C

1.) (x²+3)²–4(x²+3)–5=0

Use PEMDAS to give

x⁴ + 2x² - 8 = 0

Factor the quadratic

(x² + 4)(x² - 2) = 0

Solve for x

(x² + 4) = 0

x = ±√-4 Not a real solution since it yields the square root of a negative number ±2i an imaginary number

(x² - 2) = 0

x = ±√2 This is a real solution

You can graph the equation above at Desmos.com to confirm the x intercepts ±1.414 and see the vertex etc.

If you would clarify number 2 and as to whether it says

x/2x or x/(2x -3) inside the Absolute Value enclosures we can help you with that too.

If it is ¦x/(2x-3)¦≥1 the solution range is 1≤x< 3/2 or 3/2<x≤3 which you can also graph at Desmos.com

Please get back to us.