Vtyjkyuk

The original equation is $2sqrt x=x-2$ and I replaced $x$ with $4-2sqrt3$. I am not sure what I did wrong with the algebra. Could someone please help me. My work is posted below.

The problem is on the very first line:

$$2sqrt4-2sqrt3 stackrel?= 4-2sqrt3-2.$$

The quantity on the left side of the equation is positive, but the quantity on the right is negative.
Therefore it is impossible for this equality to be true.

What the rest of the steps show is that the left-hand side and right-hand side are exactly opposite. (Different signs but same magnitude.)
That is,
$$2sqrt4-2sqrt3 = -(4-2sqrt3-2).$$

There are a couple of things that would be marked by your teacher.

It is improper to write

$$ left(2sqrt4-2sqrt3=2-2sqrt3right)^2 $$

This is not how one squares both sides of an equation. The proper way to square both sides of an equation is

$$ left(2sqrt4-2sqrt3right)^2=(2-2sqrt3)^2 $$

But that is not the primary error.

Presumably you are trying to show that $sqrt4-2sqrt3$ is a solution to the equation $2sqrt x=x-2$. However, what you have done is not the proper way to do that. In your very first step, you have assumed the truth of what you want to demonstrate when you set the two expressions equal to each other.

The correct ways to verify that the number is a correct solution are

1. Actually solve the equation


2. Simplify the value of the expression on one side until it equals the value of the expression on the other.


3. Simplify separately the values of the expressions on each side until they are clearly the same value.

I suppose that when you make the substitution $x=4-2sqrt3$ in $2sqrtx=x-2$ you expect to end with something like $$textLHSne textRHS.$$ But this was not the case. And your algebra computations are correct. So, what is the problem? It only can be with the equality

$$2sqrt4-2sqrt3=4-2sqrt3-2.$$ And certainly it is because

$$2sqrt4-2sqrt3>0>4-2sqrt3-2.$$

And the reason to get $textLHS= textRHS$ is squaring. Note that $-1ne 1$ but $(-1)^2=1^2.$

your calculations are correct However if you are looking for how to solve this equation
put $t=sqrtx$

your equation becomes
$$2t=t^2-2$$
$$t^2-2t-2$$
$$t=1pmsqrt3$$
but $sqrtx$ is positive thus only $1+sqrt3$ is valid
$$sqrtx=1+sqrt3$$
$$x=1+3+2sqrt3$$
$$x=4+2sqrt3$$

$$2sqrt4-2sqrt3=2sqrt3-2sqrt3+1=2sqrt(sqrt3-1)^2=2|sqrt3-1|=2sqrt3-2neq2-2sqrt3$$

Actually, $x^2=y^2Leftrightarrow x=y$ iff $xygeq0.$

Try this:

2(sqrt(x))=2-x

Square the 2 in the first term, put it under the radical with the x

Gives:
Sqrt (4x)=2-x

Square both sides

4x= (2-x)^2

4x= x^2-4x+4

0=x^2-8x+4

Quadratic equation: ((-b^2(+ ; -) sqrt (-b^2 -4ac)) / (2a))

For ax^2+ bx + c

Which is x^2-8x+4

So that gives us:

((8+sqrt(-8^2 -4(1)(4))) / (2(1)))

And

((8-sqrt(-8^2 -4(1)(4))) / (2(1)))

Which equals:

((8+sqrt(64-16)) / (2))

And

((8-sqrt(64-16)) / (2))

Equals

((8+sqrt(48)) / (2))

((8-sqrt(48)) / (2))

Equals

((8+4sqrt(3)) / (2))

((8-4sqrt(3)) / (2))

Simplify:

(8/2) + ((4sqrt(3))/2)

(8/2) - ((4sqrt(3))/2)

=
4+2sqrt(3)

4-2sqrt(3)

You can eliminate the extraneous 4-2sqrt(3)

Leaving you with 4+2sqrt(3)

Good luck.

Nothing is wrong (except I never saw squaring of equations like that).

You calculations show, that $x = 4 - 2sqrt3$ is indeed the solution of your original equation.

Try other numbers. There is one more.