Question

What are the four points of intersection between 4x^2 + y^2 - 4y - 32 = 0 and x^2 - y - 7 = 0 ? Solve algebraically

Answer 1

Remember (a²-b²)=(a-b)(a+b)

solve for  a single variable
solve for y in 2nd

add y to both sides
x²-7=y
sub (x²-7) for y in other equaiton

4x²+(x²-7)²-4(x²-7)-32=0
expand
4x²+x⁴-14x²+49-4x²+28-32=0
x⁴-14x²+45=0
factor
(x²-9)(x²-5)=0
(x-3)(x+3)(x-√5)(x+√5)=0
set each to zero

x-3=0
x=3

x+3=0
x=-3

x-√5=0
x=√5

x+√5=0
x=-√5


sub back to find y

(x²-7)=y

for x=3
9-7=2
(3,2)

for x=-3
9-7=2
(-3,2)

for √5
5-7=-2
(√5,-2)

for -√5
5-7=-2
(-√5,-2)


the intersection points are

(3,2)
(-3,2)
(√5,-2)
(-√5,-2)