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WARRIOR [948]
3 years ago
7

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

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0
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)
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Step-by-step explanation:

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What’s the pattern of -1, -8, -27, and -64
AURORKA [14]

Hello from MrBillDoesMath!

Answer:

a(n) = (-n)^3  where n = 1,2,3,...


Discussion:

The pattern  1,8,27, 64... is immediately recognizable as the the cube of the positive integers. But this question has a minus sign appearing before  each entry, suggesting we try this:


 - 1  =  (-1)^3

 -8  = (-2)^3

-27  = (-3)^3

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That's what the problem statement asked for . The answer is equivalently

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Thank you,

MrB

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