Question

What is the solution of x = 2 + StartRoot x minus 2 EndRoot?

Answer 1

Answer:

x = 2 or x = 3

Step-by-step explanation:

we have the expression:

$x=2+\sqrt{x-2}$

First we must remove the square root, for this we pass the 2 to the left:

$x-2=\sqrt{x-2}$

and we move the square root to the left as an exponent of two:

$(x-2)^2=x-2$

We develop the square binomial:

$x^2-4x+4=x-2$

and clear the right side for the equation to be equal to zero:

$x^2-4x+4-x+2=0$

combining like terms:

$x^2-5x+6=0$

Because it is a quadratic equation we will have two answers, which we can find by the general formula or by factoring:

To factor we open two parenthesis, both with an x, and look for two numbers that when multiplied result in +6 and when added result in -5, those numbers are -2 and -3, so the factoring is:

$(x-2)(x-3)=0$

And now we use the Zero Product Property, whic is if two thing are multiplied and result in zero one of the two or both things must zero, in this case:

$(x-2)=0\\x=2$

and

$(x-3)=0\\x=3$

the solutions are: x = 2 or x = 3; both satisfy the equation.

Answer 2

Answer:

Option C on E2020

Step-by-step explanation:

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