Question

can anyone explain how to do this? determine all of the real number solutions. x-sqrtx=20

Answer 1

Use u-substitution:
let:
\[u=\sqrt{x}\]
Rewrite:
\[u^2-u-20=0\]

Answer 2

Factor

Answer 3

why u\[u^{2?}\]

Answer 4

You will see, factor that equation above

Answer 5

\[u^{2}-u-20=0\]

Answer 6

yes, factor it

Answer 7

\[u^{2}-u=20 ?\]

Answer 8

u=20

Answer 9

\[(u-5)(u+4)=0\]

Answer 10

right?

Answer 11

yes

Answer 12

ok now that we know this, we can substitute sqrt(x) back in for u

Answer 13

\[u-5=0, \sqrt{x}-5=0, \sqrt{x}=5\]
\[(\sqrt{x})^2=5^2, x=25\]

Answer 14

16 is not a solution

Answer 15

there are two roots one real one complex this cannot be

Answer 16

There is only one root.

Answer 17

I was able to find only 1 root which is 25 correct

Answer 18

yes

Answer 19

ok

Answer 20

Thanks

Answer 21

np

Answer 22

\[u+4=0\]
\[\sqrt{x}+4=0\]
\[\sqrt{x}=-4\]
we know that \[if \sqrt{x }\qquad x \ge 0 \qquad always\]

Answer 23

Yes but you can't simply square both sides here, check 16 in the original equation, it doesn't work

Answer 24

No solutions exist for
\[\sqrt{x}=-4\]

Answer 25

I know what you mean..
I can check it in original eq.