Can a suitable substitution help?

$x+\sqrt{x-2}=4$ Squring & solving, $\Rightarrow x^2-9x+18=0$ $\Rightarrow x=6,3$ Due to squaring, there's an extraneous root of the final quadratic. Hence one should check the roots in the original equation. Here $x=6$ fails the test hence we're left with $x=3$ as the only solution.

Since the question asks about the roots so we proceed to find it:

$\sqrt { x-2 } =4-x\\ or,\quad x-2=16-8x+{ x }^{ 2 }\\ or,\quad { x }^{ 2 }-9x+18=(x-6)(x-3)=0.\\ \therefore x=6\quad or\quad 3$

Now, we get two real value of x for the equation ${ x }^{ 2 }-9x+18=0$ and not for $\sqrt { x-2 } =4-x$ .

So we crosscheck, out of the two value of x obtained for ${ x }^{ 2 }-9x+18=0$ does both satisfy $\sqrt { x-2 } =4-x$ .

Then we find that only x=3 satisfies $\sqrt { x-2 } =4-x$ and not x=6 so the answer is only 1 real root.

Notice that $\sqrt { x-2 } =4-x$ and ${ x }^{ 2 }-9x+18=0$ are not equivalent

Everybody does this : sqrt(x-2) = 4-x ; x-2 = (4-x)^2; x^2-9x+18 = 0; x = 3,6; i.e. two real roots.

But, You must always check the Domain before solving any eqn.

Here, LHS is an always positive value (since, square root.) So, RHS > 0, or, 4-x > 0, x<4. 3 satisfies this, but, 6 does not.

no real solution is possible if x is less than 2 because the square root of a negative number does not exist within the real numbers.

obviously, no solution is possible, for real x values greater than 4.

So, real solutions are possible only for x values between 2 and 4.

it follows that the only real solution is x = 3.

there could be 2 roots, since 6 minus square root of (6-2) is equal to 6 minus the square root of 4, since there are two roots to square root of any positive numbers, the square root of 4 is equal to +2 and -2, hence 6 + (-2) = 4

$x + \sqrt(x-2) - 4 = 0$

we can factorize it directly

$(\sqrt(x-2) -2)(\sqrt(x-2) + 1 )$

and here we get

$x = 3$ and $x=6$ but that root is extraneous

so $x = 3$

x+√(x-2)=4

x+√(x-2)-4=0

(x-2)+√(x-2)-2=0

let y=√(x-2)

y^2+y-2=0

(y+2)(y-1)=0

Therefore

y=-2

y=1

Furthermore

y=√(x-2)=-2

((√(x-2))^2=(-2)^2

x-2=4

x=6

y=√(x-2)=1

(√(x-2))^2=(1)^2

x-2=1

x=3

Substituting it to the original equation tell you that 6 is not an acceptable answer.

Therefore the answer is x=3.

How is this problem level 3, honestly? In my honors trig class last year, these were like the "A" level problems (easiest) and this is America

$x+\sqrt{x-2}=4(x>=2)$ $\Leftrightarrow(x-2)+\sqrt{x-2}-2=0$ $\Leftrightarrow(\sqrt{x-2}-1)(\sqrt{x-2}+2)=0$ $\Leftrightarrow\sqrt{x-2}-1=0$ $\Leftrightarrow\sqrt{x-2}=1$ $\Leftrightarrow x=3$

Oh C'mon. Just substitute 3 into the equation, and you get 4.
So the 3' is the lone real root

Subtract 4 from both sides and get the function: f(x) = x + sqr(x-2) - 4

Once we graph the function, we see that there is only one x-intercept.

Hence, 1 real root.

Note that the function is always increasing. Therefore, the function can only be equal to 4 at one x-value, which is 3, by inspection.