√(4x+1)=5-x?

Question

1. √(4x+1)=5-x

2. 2x^-4 – 5x^-2 + 2 = 0

3. 3x^2/-3 + 2x^1/-3 - 8 = 0

Answer 1

sqrt(4x+1)=5-x
4x+1>=0 s0 x>=-1/4 and 5-x>=0 so x<=5
Now you can square both sides
4x+1= 25 -10x+x^2
x^2-14x+24=0 x=((14+-10))/2 = 12 and x= 2 so x= 2 as x=12 is not a solution

2) Multiply both sides by x^4
so2x^4-5x^2+2=0
If you call x^2 = z you get 2z^2-5z+2=0

z=((5+-sqrt(25-16)/4 so z= 2 and z= 1/2
x=+-sqrt2 and x= +-1/2sqrt2

3) multiply by -3
3x^2+2x+24=0 No real roots