The problem I have is:
x + √3+√x =3
*keep in mind 3+ √x is inside the square root.
Now logically, I would do this:
√3+√x =3-x /2
3+√x= (3-x)2
√x=(3-x)2 - 3
Square that too and calculate everything. However, It gets very complicated like that and these problems are supposed to be quick solving. I looked at the solution and I do not understand it at all.
It says that the problem has two cases:
1st case x > 1 so it's
x + √3 + √ x > 3
2nd case 0 ≤ x < 1 so it's
x + √3 + √ x < 3
How did we get the inequality now and why?
Thank you in advance
x + √3+√x =3
*keep in mind 3+ √x is inside the square root.
Now logically, I would do this:
√3+√x =3-x /2
3+√x= (3-x)2
√x=(3-x)2 - 3
Square that too and calculate everything. However, It gets very complicated like that and these problems are supposed to be quick solving. I looked at the solution and I do not understand it at all.
It says that the problem has two cases:
1st case x > 1 so it's
x + √3 + √ x > 3
2nd case 0 ≤ x < 1 so it's
x + √3 + √ x < 3
How did we get the inequality now and why?
Thank you in advance