Secondary 2 Maths
1 Answer
We have
\[500=(5x-5y)^2=(5x)^2+(5y)^2 -2\cdot (5 x)(5y)=25x^2+25y^2-50xy.\]
Since $xy=5$,
\[500=25x^2+25y^2-50xy=25x^2+25y^2-50\cdot 5=25x^2+25y^2-250\]
\[\Rightarrow 500+250=25x^2+25y^2\]
\[\Rightarrow 750=25(x^2+y^2)\]
\[\Rightarrow x^2+y^2=\frac{750}{25}=30.\]
Savionf
573
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 270 views
- Pro Bono
Related Questions
- Abstract Algebra : Commutativity and Abelian property in Groups and Rings
- Does $\lim_{(x,y)\rightarrow (0,0)}\frac{(x^2-y^2) \cos (x+y)}{x^2+y^2}$ exists?
- Let $z = f(x − y)$. Show that $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=0$
- Solve this problem using branch and bound algorithm.
- Fields and Galois theory
- Solving Inequalities- Erik and Nita are playing a game with numbers
- Show that the $5\times 5$ matrix is not invertable
- Is the $\mathbb{C}$-algebra $Fun(X,\mathbb{C})$ semi-simple?
