Question

Solve for y.
y*(sqrt(1-y^2)) = 0.5

Answer 1

This is the equation:
\[y \sqrt{1-y ^{2}}= 0.5\]

Answer 2

try squaring both sides...what do you get?

Answer 3

is it:
\[y^{2} (1-y^{2}) = 0.25\]

Answer 4

yes. now distribute y^2 into 1 - y^2

Answer 5

y^2 - y^4 = 0.25

Answer 6

right. now subtract 0.25 from both sides

Answer 7

so you get
y^2 - y^4 - 0.25 = 0
and i think you have to make all the terms have the same denominatot

Answer 8

\[\frac{ 4y - 4y^2 -1 }{ 4 }\]

Answer 9

aren't you forgetting = 0?

Answer 10

\[\frac{4y - 4y^2 - 1}4 = 0\]
now...if you cross multiply..
\[4y - 4y^2 - 1 = 0\]
yes?

Answer 11

yeuupp now what?

Answer 12

wait... why are you using y^2? it should be \[4y^2 - 4y^4 - 1 = 0\]

Answer 13

now multiply by -1... \[4y^4 - 4y^2 + 1 = 0\]
right?

Answer 14

yes i did that
and sorry typo :O
Well now do we move 1 to the other side, so we subtract both sides by one?

Answer 15

we just moved 1 to this side..why move it back?

Answer 16

true, go on.

Answer 17

now.. you perform quadratics.. let \[a = y^2\]
so this becomes\[4a^2 - 4a + 1 = 0\]
solve for a

Answer 18

do you know how to?

Answer 19

yes, give me a sec.

Answer 20

sure

Answer 21

(2a - 1)^2 = 0

Answer 22

right. now solve for a

Answer 23

does a = 1/2 or \[a = \pm \frac{ 1 }{ 2 }\]

Answer 24

not \(\pm\)...just +1/2

Answer 25

so you have \[a = \frac 12\]
then remember i said a = y^2

\[y^2 = \frac 12\]
solve for y

Answer 26

\[y = \sqrt{\frac{ 1 }{ 2 }}\]

Answer 27

close... \[y = \pm \sqrt{\frac 12}\]
ironic how you didn't use that \(\pm\) when you should have

Answer 28

aha thankss man
Since this is part of a system of equations, do i sub in the negative an positive values?

Answer 29

what do you mean?

Answer 30

substitute y = sqrt (1/2)
substitute y = -sqrt(1/2)

Answer 31

i meant part of systems of equations and substituting....oh wait i know....anyway..yes you sub

Answer 32

alriight thanks man