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 $xy=5$,
\[500=25x^2+25y^2-50xy=25x^2+25y^2-50\cdot 5=25x^2+25y^2-250\]
\[\Rightarrow 500+250=25x^2+25y^2\]
\[\Rightarrow 750=25(x^2+y^2)\]
\[\Rightarrow x^2+y^2=\frac{750}{25}=30.\]

574
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 347 views
- Pro Bono
Related Questions
- When is Galois extension over intersection of subfields finite
- Attempting to make a formula/algorithm based on weighted averages to find how much equipment we need to maintain.
- Induced and restricted representation
- Algebra Question
- Find $x$ so that $\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible
- Population Equations
- Does $\lim_{(x,y)\rightarrow (0,0)}\frac{(x^2-y^2) \cos (x+y)}{x^2+y^2}$ exists?
- How to parameterize an equation with 3 variables