How do you solve #x^2 + y^2 = 4# and #y^2 = 3x#?

2 Answers
Aug 4, 2015

Answer:

#(x,y)=(1,-sqrt(3)),(1,sqrt(3)),(4,isqrt(12)),(4,-isqrt(12))#

Explanation:

Substitute the second equation into the first to obtain a quadratic equation for #x#:

#x^2+y^2=x^2+3x=4# => #x^2+3x-4=(x+4)(x-1)=0#

This has solutions #x=-4,1#, substituting this into the second equation we have #y=+-sqrt(3),+-isqrt(12)#.

Therefore we have:

#(x,y)=(1,-sqrt(3)),(1,sqrt(3)),(4,isqrt(12)),(4,-isqrt(12))#

Aug 4, 2015

Answer:

Substitute the second equation into the first to get a quadratic in #x#, the positive root of which gives two possible Real values for #y# in the second equation.

#(x, y) = (1, +-sqrt(3))#

Explanation:

Substitute #y^2=3x# into the first equation to get:

#x^2+3x = 4#

Subtract #4# from both sides to get:

#0 = x^2+3x-4 = (x+4)(x-1)#

So #x = 1# or #x = -4#.

If #x = -4# then the second equation becomes #y^2 = -12#, which has no Real valued solutions.

If #x = 1# then the second equation becomes #y^2 = 3#, so #y = +-sqrt(3)#