Question

Can you create an equation for me with an extraneous solution, please? Thanks!

Answer 1

you mean like one with a square root?

Answer 2

Yes please

Answer 3

Like sqrt (x - 98) + 2 = 1

Answer 4

ok lets make it easy
pick a number you want to be a solution

Answer 5

Okay, 4?

Answer 6

ok so \(\sqrt{4+5}=3\) right? so we will make one side of the equation
\[\sqrt{x+5}\] and we know if \(x\) is \(4\) then that side will be \(3\)

Answer 7

now for the other side
\[\sqrt{4+12}=4\] so on the other side we can put \(\sqrt{x+12}-1\) which will also be \(3\)
now lets solve
\[\sqrt{x+1}=\sqrt{x+12}-1\]

Answer 8

typo there i meant
lets solve
\[\sqrt{x+5}=\sqrt{x+12}-1\]

Answer 9

oh damn i think i messed this one up, will only get one solution

Answer 10

Oh lol, well would the first step be to square both sides we get x + 5 = x + 12 - 1?

Answer 11

no, when you square both sides you get
\[x+5=x+12-2\sqrt{x+12}+1\]

Answer 12

lets try this one, might work better
if \(x=7\) then \(\sqrt{2x+2}=\sqrt{16}=4\) and \(\sqrt{x+2}=3\) so lets try and solve
\[\sqrt{2x+2}=\sqrt{x+2}+1\]

Answer 13

Wow, I'm sorry, I really suck at math and I'm lost

Answer 14

ok let me see if i can come up with a really easy example

Answer 15

Thank you! I really appreciate it.

Answer 16

\[x-6=\sqrt{x}\]

Answer 17

much easier