(2x+1)^1/2=(5-2x)^1/2 help..............?

Question

trouble on it

Answer 1

Square both sides:

2x+1 = 5 - 2x ---> 4x = 4 ---> x=1

Answer 2

a function raised to the 1/2 is the square root of that function. in this case, both functions are square root. to solve this, first you must eliminate the exponents (in this case, square roots) by raising both sides to theit least common exponent (in this case by two or "square") and multiplying the exponents. thus,

[(2x + 1)^1/2]^2 = [(5 - 2x)^1/2]^2

wherein 2 is the least common exponent. then, multiplying the exponents 1/2 x 2, this will result to

(2x+1) = (5 - 2x)

simplifying the equation, you get

2x + 2x = 5 - 1

where you get

4x = 4 or x = 1

in cases where in one side of the equation has exponent 1/3 and the other is 1/2, you rasie both sides by 6. ex. x^1/3 = y^1/2 (where x and y are equations), raise bothe sides by 6 and you'll get x^2 = y^3 then you can simplify from there.

Answer 3

* is square root
=> *(2x+1)=*(5-2x)
<=> 2x+1=5-2x
and 2x+1>0
and 5-2x>0
<=> x=1
and x>-1/2
and x<2.5
=> x=1

Answer 4

(2x+1)^1/2=(5-2x)^1/2
square both sides of the equation to take off square root...
which leaves us to..
2x+1=5-2x
transpose all with variables in left side and numbers on right...
2x+2x=5-1
compute...
4x=4
x=1

Answer 5

(2x+1)^1/2=(5-2x)^1/2
2(2x+1)=2(5-2x)
4x+2=10-4x
4x+4x=10-2
8x=8
x=8/8
x=1