Finding Possible Solutions Part 2?

Question

1. x^4 - 5x^3 + 5x^2 + 5x = 6

2. 2x^3 - 5x^2 - 6x + 4 = 0

3. 2^3x = 1/256

4. x - 1/2 - 3x - 4/6 = 1/3

5. 6x - 1 = 15/x

6. absolute value x^2 - 5 = 4

Thanks...

Answer 1

1.x^4 - 5x^3 + 5x^2 + 5x = 6

<=> x^4 - (5x^3 + 5x) + 5x^2 - 5 - 1=0

<=> (x^4-1) -5x(x^2-1) + 5(x^2-1)=0

<=> (x^2-1)(x^2+1) -5x(x^2-1) +5(x^2-1)=0

<=>(x^2-1)( x^2+1 -5x +5)=0

<=>(x-1)(x+1)(x^2-5x+6)=0

<=>x=1 or x=-1 or x=3 or x=2

2. (no idea! )

3.2^3x = 1/256

<=>2^(3x)=1/(2^8)

<=>2^8 * 3^(3x)=1

<=>8+3x=0

<=>x=-8/3

4. x - 1/2 - 3x - 4/6 = 1/3

<=> -2x-3/2=0

<=>-2x=3/2

<=>x=-3/4

5.  6x - 1 = 15/x    (x<>0)

<=> 6x^2-x-15=0 (x<>0)

<=> x=-3/2 or x=5/3

6. |x^2 - 5| = 4

<=> x^2-5=4 where x>=2 or x<= -2 (1)

and 5-x^2=4 where -2<x<2     (2)

^2

(1) <=>x=9 <=>x=3 or x=-3

(2) <=> x^2=1 <=> x=1 or x=-1

So, the equation has four roots: x=3 or x=-3 or x=1 or x=-1