Question

The solution to x2 − 12x = 0 is {1, 12} true or false?

Answer 1

false

Answer 2

So we have an ordered pair

(1 , 12) as we know that is (x , y)

so if we plug in 1 for 'x' in your equation do we get the result that y= 12?

lets check

\[\large x^2 - 12x = 0\]
\[\large (1)^2 - 12(1) = 0\]
\[\large 1 - 12 = 0\]
\[\large -11 \cancel{=} 0\]

so it is false

Answer 3

@johnweldon1993 .(0,12) is this a solution to the above question?

Answer 4

No, there are only 2 solutions to this equation and they are

\[\large x^2 - 12x = 0\]
\[\large x^2 - 12x + 36 = 36\]
\[\large (x - 6)^2 = 36\]
\[\large x - 6 = \pm6\]
\(\large x = 0\) or \(\large x = 12\)

Answer 5

Same thing.
(0,12) is the solution.
x1 = 0 and x2 = 12

Answer 6

Or in ordered pair form

(0 , 0) or (12 , 0)

as you can see 'y' will always be 0 because that is the solution to a polynomial (where the parabola crosses the x-axis)

Answer 7

yes , we have no y terms in our equation so y will be always zero. Thanks.