Question

Solve for y: y+y^2=x^2+c

Answer 1

@amistre64

Answer 2

you have a quadratic in y

Answer 3

y^2 + y - (x^2 + c) = 0

Answer 4

And then?

Answer 5

and then use the quadratic formula ...

Answer 6

Let me work it out.

Answer 7

ill let you :)

Answer 8

I got y=-1+/-sqrt(4x^2+4c+1) divided by 2 but if I plug in 2 into x, I got c=-4. The answer for this problem should be y=-1+sqrt(4x^2-15) divided by 2.

Answer 9

It's an initial-value problem where y(2)=0.

Answer 10

\[y^2 + y - (x^2 + c) = 0\]
\[y=\frac{-1\pm\sqrt{1^2+4(x^2+c)}}{2}\]
\[y=\frac{-1\pm\sqrt{4x^2+c}}{2}\]

Answer 11

im assuming c is an all encompassing constant

Answer 12

But why doesn't the answer doesn't contain +/- and only + sign?

Answer 13

\[0=\frac{-1\pm\sqrt{16+c}}{2}\]
\[0=-1\pm\sqrt{16+c}\]
\[1=\sqrt{16+c}\]
c = -15

Answer 14

its just a matter of the mathing process. 1 is a postive number, and there is no sqrt value that is negative

Answer 15

no 'real' sqrt ...

Answer 16

So what's the correct answer? The one that doesn't contain +/- sign and only the plus sign?

Answer 17

the +- sign is an operation that allows us to + or - the sqrt as needed.

it just so happens that in this case there is no subtraction operation that produces a valid real solution

Answer 18

\[1 = \pm\sqrt{16+c}\]
\[1=\sqrt{16+c}~~or~~1=-\sqrt{16+c}\]
since there is not sqrt(n) that gives us a negative value, the right option is simply not valid

Answer 19

Thanks!

Answer 20

yw

Answer 21

:)