Question

1+x^2+y^2=13 using quadratic equation

Answer 1

what to find ?

Answer 2

\[1 + x^{2} + y^{2} = 13 \implies y = \pm \sqrt{12-x^{2}}\] Assuming you mean that we are supposed to use the quadratic formula to solve. If you do mean that, then wehave to solve for y like so. Now, since we want to set the equation = to 0 and solve for x, we have to realize that the plus or minus on the outside NOR the square root matter. If we find x such that the function is 0, the function will still be 0 regardless of whether or not that root or plus or minus is there. So we can simply worry about solving this:
\[y = 12 - x^{2}\] Using quadratic formula from there would give us:
\[\frac{ 0 \pm \sqrt{0^{2} - 4(-1)(12)} }{ -2 } \implies \pm\frac{ \sqrt{48} }{ -2 } \implies \pm\frac{ 4\sqrt{3} }{ -2 }\implies \pm -2\sqrt{3}\]. So you get thw two answers of

x = 2sqrt(3)
x = -2sqrt(3)

Answer 3

\[+2\sqrt{3} , -2\sqrt{3}\]