Can you help me solve this radical equation? (2x - 1)^1/2 = x - 2. I am out of my element w/algebra. Thanks!

Answer 1

(2x - 1)^1/2 = x - 2.
squaring both sides we get
2x-1 = (x-2)^2
2x- 1 = x^2 +4- 4x
x^2-6x +5 =0
(x-1)(x-5)=0

so x = 1 or 5

Answer 2

When something is raised to a fractional power, you raise it to the numerator and the take the denominator root. So, anything to the power of 1/2 would be raised to hte power of 1 (same number), then the 2nd root. If it was to the power of 2/3, for example, you would raise to the 2nd power, then take the 3rd root. See?

So,

(2x - 1)^1/2 = x - 2
root(2x - 1) = x - 2

See?

Square both sides.

2x - 1 = (x - 2)(x - 2)
2x - 1 = x^2 - 4x + 4
0 = x^2 - 6x + 5
0 = (x - 1)(x - 5)
x = 1, 5

~_~_~_~_~_~_~_~_~_~

Another way to look at it, is this. When you raise a power to a power, you multiply the exponents right? You can try to get the exponent to equal the power of 1. You know that 2/2 = 1 and 3/3 = 1, so what you need to do, is raise the exponent to an exponent that will get a whole number power. This one, you would square it.

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

Solved the same as above from this step.

You could also raise it to a power that is divisible by 2, but that would make it longer. For example, raise to the power of 4.

[(2x - 1)^1/2]^4 = (x - 2)^4
(2x - 1)^4/2 = (x - 2)^4
(2x - 1)^2 = (x - 2)^4

You could solve from there. But of course, anytime there is a 2 in the denominator of an exponent, just square it. You will have smaller numbers that way.

Answer 3

(2x - 1)^(.5) = x - 2

square both sides
2x - 1 = x^2 - 4x + 4

subtract 2x and add 1 for both sides
0 = x^2 - 6x + 5

factor
0 = (x - 5) (x - 1)

x = 5 or 1

You need to check your answer everytime you solve a root equation

(2*1 - 1)^(.5) = 1 - 2
(2 - 1)^(.5) = -1
1^(.5) = -1
1 = -1

1 is not equal to -1, thus, 1 is not the solution

(2*5 - 1)^(.5) = 5 - 2
(10 - 1)^(.5) = 3
(9)^(.5) = 3
3 = 3

the statement is true

The only solution to this problem is 5

Answer 4

2x - 1 = x^2 - 4x + 4
0 = x^2 - 6x + 5
0 = (x - 1) (x - 5)
x = 1 or 5

check answer: 1 is not the answer because when you plug it back in, -1 is not equal to 1.

ans: x = 5

Answer 5

Square both sides:

(2x - 1) = (x - 2)^2

2x - 1 = x^2 - 4x + 4

x^2 - 6x + 5 = 0

You can solve this with the quadractic formula, or you can look at it and see that it factors to

(x - 1)(x - 5) = 0

So x = 1 or 5

If you use 1, you get 1^(1/2) = -1, which is true.

If you use 5, you get 9^(1/2) = 3, which is also true.

Answer 6

2x - 1 = x² - 4x + 4
x² - 6x + 5 = 0
(x - 5).(x - 1) = 0
x = 1 , x = 5