Can someone help me solve and check

Question

sqrt(2x-3)=3-x

Answer 1

√(2x-3) = 3-x

 

It's hard to solve when there's a radical sign, so the first thing the you need to do is to remove it. You can do this by raising both sides of the equation by the power of 2:

 

2x - 3 = (3 - x)²

 

Expanding the right side:

 

2x - 3 = 9 - 6x + x²

 

Transferring everything on any side that you prefer, this time the right side:

 

0 = x² - 8x + 12

 

So by factoring, you will get:

 

(x-6)(x-2) = 0

 

x - 6 = 0, x - 2 = 0

 

x = 6, x = 2

 

By checking, just substitute them in the given equation:

 

Check x = 6 first:

 

sqrt(2(6)-3) = 3 - 6

 

sqrt(9) = -3

 

It is correct because the answer in the square root is always + and/or -.

 

Next x = 2:

 

sqrt(2(2)-3) = 3 - 2

sqrt(1) = 1

 

1 = 1

 

In this problem, the answers are: x = 2 and x = 6. *READ THE REMINDER BELOW*

 

*REMINDER: When solving equations with square root, you should always check the answers that you get, because sometimes, even the process it correct, it will yield a negative value inside the square root (ex. √(-3)), that will make it imaginary. We call this solutions, ERRONEOUS solution and should be EXCLUDED in your final answer.*

Answer 2

sqrt(2x-3)=3-x

 

-> 2x-3 = (3-x)^2

 

->2x-3 = 9 - 6x - x^2

 

-> x^2 + 6x + 2x -3 -9 = 0

 

-> x^2 +8x -12 = 0

 

Using quadratic equation- a= 1, b= 8, c= -12

 

x= (-b(+/-) sqrt(b^2 - 4ac))/ 2a

 

 

x= (-8(+/-) sqrt((-8)²- 4(1)(-12)))/ 2(1)

 

 

Plugging this into a calculator-> x=1.29, -9.29