What is the solution of x=2+ square root x-2? a. x = 2 b. x = 3 c. x = 2 or x = 3 d. no solution

So I'm going to assume that this question is asking for non extraneous solutions, or solutions that are found in the equation and are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

Next, square both sides:

Next, subtract x and add 2 to both sides of the equation:

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

Now you can rewrite the equation as

Now, apply the Zero Product Property and solve for x as such:

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

Since both solutions hold true when x = 2 and x = 3, your answer is C. x = 2 or x = 3.

its not b

step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.