How do you find the exact solutions to the system #y+x^2=3# and #x^2+4y^2=36#?
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The solutions are #(0,3)# and #(+-sqrt(23)/2, -11/4)#
#y+x^2=3#
Solve for y:
#y=3-x^2#
Substitute #y# into #x^2+4y^2=36#
#x^2+4(3-x^2)^2=36#
Write as the product of two binomials.
#x^2+4(3-x^2)(3-x^2)=36color(white)(aaa)#
#x^2+4(9-6x^2+x^4)=36color(white)(aaa)#Multiply the binomials
#x^2+36-24x^2+4x^4=36color(white)(aaa)#Distribute the 4
#4x^4-23x^2=0color(white)(aaa)#Combine like terms
#x^2(4x^2-23)=0color(white)(aaa)#Factor out an #x^2#
#x^2=0# and #4x^2-23=0color(white)(aaa)#Set each factor equal to zero
#x^2=0# and #4x^2=23#
#x=0# and #x=+-sqrt(23)/2color(white)(aaa)#Square root each side.
Find the corresponding #y# for each #x# using #y=3-x^2#
#y=3-0=3, and, y=3-23/4=-11/4#
Hence, the solutions are, #(1) x=0, y=3; (2 and 3) x=+-sqrt23/2, y=-11/4#.
Note that there are three solutions, which means there are three points of intersection between the parabola #y+x^2=3# and the ellipse #x^2+4y^2=36#. See the graph below.
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Three points of intersection #(-sqrt(23)/2, -11/4)#, #(sqrt(23)/2, -11/4)# and #(0, 3)#
Given:
#y + x^2 = 3#
#x^2 + 4y^2 = 36#
Subtract the first equation from the second:
#4y^2 - y = 33#
Subtract 33 from both sides:
#4y^2 - y - 33 = 0#
Compute the discriminant:
#b^2 - 4(a)(c) = (-1)^2 - 4(4)(-33) = 529#
Use the quadratic formula:
#y = (1 + sqrt(529))/8 = 3# and #y = (1 - sqrt(529))/8 = -11/4#
For #y = 3#:
#x^2 = 3 - 3#
#x = 0#
For #y = -11/4#:
#x^2 = 3 + 11/4 #
#x^2 = 12/4 + 11/4 #
#x^2 = 23/4#
#x = sqrt(23)/2# and #x = -sqrt(23)/2#
