Does Graphing Work Here?

Algebra Level 2

Given that:

{ x 4 + x 2 y 2 + y 4 = 700 x 2 + x y + y 2 = 35 \begin{cases} x^4+x^2y^2+y^4= 700 \\ x^2+xy+y^2= 35\end{cases}

What is 2 x y 2xy to the nearest whole number?

25 20 15 30 35

Hana Wehbi by Hana Wehbi , 51, USA

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3 solutions

Hana Wehbi
Feb 7, 2021

x 4 + x 2 y 2 + y 4 = ( x 2 + x y + y 2 ) ⏟ 35 ( x 2 − x y + y 2 ) = 700 ⟹ x 2 − x y + y 2 = 700 35 = 20 x^4+x^2y^2+y^4= \underbrace{(x^2+xy+y^2)}_{35}(x^2-xy+y^2) =700\implies x^2-xy+y^2 =\frac{700}{35}= 20 .

{ x 2 + x y + y 2 = 35 − x 2 − x y + y 2 = 20 \begin{cases} x^2+xy+y^2 = 35 \\ - \\\ x^2-xy+y^2 = 20 \end{cases}

⟹ 2 x y = 35 − 20 = 15 \implies 2xy = 35-20=15

2 Helpful 0 Interesting 2 Brilliant 0 Confused

Did you want us to graph this quartic function?

Jason Gomez - 4 months ago

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It’s another optional solution.

Hana Wehbi - 4 months ago
Chris Lewis
Feb 9, 2021

( x 2 + x y + y 2 ) 2 − ( x 4 + x 2 y 2 + y 4 ) = ( x 4 + 2 x 3 y + 3 x 2 y 2 + 2 x y 3 + y 4 ) − ( x 4 + x 2 y 2 + y 4 ) = 2 x 3 y + 2 x 2 y 2 + 2 x y 3 = 2 x y ( x 2 + x y + y 2 ) \begin{aligned} {\color{#D61F06}\left(x^2+xy+y^2\right)^2}-{\color{#3D99F6}\left(x^4+x^2 y^2 + y^4\right)}&={\color{#D61F06}\left(x^4+2x^3 y + 3x^2 y^2 + 2xy^3 + y^4\right)}-{\color{#3D99F6}\left(x^4+x^2 y^2 + y^4\right)} \\ &=2x^3 y + 2x^2 y^2 + 2xy^3 \\ &=2xy \left(x^2+xy+y^2\right)\end{aligned}

Substituting in the given values, 3 5 2 − 700 = 2 x y ⋅ 35 {\color{#D61F06}35^2}-{\color{#3D99F6}700}=2xy \cdot 35

So 2 x y = 3 5 2 − 700 35 = 35 − 20 = 15 2xy=\frac{35^2-700}{35}=35-20=\boxed{15}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

Thank you for sharing your solution.

Hana Wehbi - 4 months ago
Iliya Hristov
May 4, 2021

( x 2 − y 2 ) 2 = ( x + y ) 2 ( x − y ) 2 = ( 35 + x y ) ( 35 − 3 x y ) = 700 − 3 x 2 y 2 ⇒ 2 x y = 15 \left(x^2-y^2\right)^2=(x+y)^2(x-y)^2=(35+xy)(35-3xy)=700-3x^2y^2\;\;\Rightarrow\;\;2xy=\boxed{15}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Bonus: Find the sum of all x x solutions of the system.
Solution: x + y = ± 35 + 15 2 + x − y = ± 35 − 3 ∗ 15 2 ⇒ x = ± 42.5 ± 12.5 2 ⇒ ∑ x = 0 x+y=\pm\sqrt{35+\frac{15}{2}}\;\;\;\;\;\;+\;\;\;\;\;\;x-y=\pm\sqrt{35-3*\frac{15}{2}}\;\;\;\Rightarrow\;\;\;x=\frac{\pm\sqrt{42.5}\pm\sqrt{12.5}}{2}\;\;\;\Rightarrow\;\;\;\sum{x}=\boxed{0}

Iliya Hristov - 1 month ago

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