+118704 Some of the other mathematicians have got much better techniques for doing these but I resorted to trial and error and just looking for patterns.
I started by adding 3 to the y value because 0 to the power of anything is 0
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | |
| y | 159 | 29 | -1 | -3 | -1 | 29 | 159 | |
| y+3 | 162 | 32 | 2 | 0 | 2 | 32 | 162 | 32=2^5 but 162 is not a power of 3 .. but 81 is |
| (y+3)/2 | 81 | 16 | 1 | 0 | 1 | 16 | 81 | |
| [(y+3)/2]^(1/4) | 3 | 2 | 1 | 0 | 1 | 2 | 3 | |
so .. working backwards
y=2x4−3
I worked this out quite quickly because I have had a lot of practice but you should get a better answer off someone else 
+2446 Ok, thanks for replying. One way to check if your solution is right is to plug it in and see if the resulting equation is indeed a true one. Let's do that!
√2x−4−x+6=0
Let's plug in x=4 and x=10 and see if we get a true statement.
Check x=4
| √2∗4−4−4+6=0 | Let's evaluate what is inside the radical to start and determine whether or not this is a true statement. |
| √8−4−4+6=0 | |
| √4−4+6=0 | The radicand, 4, is a perfect square, so this can be simplified further. |
| 2−4+6=0 | This is elementary addition and subtraction now. |
| 4=0 | This is an false. statement, so this solution does not satisfy the original. |
Check x=10
| √2∗10−4−10+6=0 | Ok, simplify inside the radical just like before determine if true just like before. |
| √20−4−10+6=0 | |
| √16−10+6=0 | Just like in the previous problem, the number underneath the square root is a perfect square, so the presence of them go away. |
| 4−10+6=0 | |
| 0=0 | This is a true statement, so this is a solution. |
Ok, I have verified that x=10 is a valid solution, but x=4 does not satisfy this equation. I can tell you that you are wrong. The only issue here is that I cannot determine if x=4 is indeed an extraneous solution or not. I'll just solve the equation.
| √2x−4−x+6=0 | Add everything but the radical expression to the right hand side of the equation. |
| √2x−4=x−6 | Square both sides to eliminate the radical. |
| 2x−4=(x−6)2 | Expand the right hand side of the equation. |
| 2x−4=x2−12x+36 | Move everything to one side of the equation again. |
| x2−14x+40=0 | This is a quadratic, so there will be another solution to this equation. Factoring is easiest here. |
| (x−10)(x−4)=0 | Now, set each factor equal to 0 and solve for x. |
| x−10=0x−4=0x=10x=4 | |
Ok, you solved the equation correctly, but x=4 is an extraneous solution. However, there is no option choice that states both that x=4 is a solution and x=10 is an extraneous one.
