Question

How many real solutions does the system $x^2+y^2=25$ and xy =10 have

• A. 4
• B. 1
• C. 2
• D. none of these

Answer 1

Answer: (A)

4

$xy=10$

$y=\frac{10}{x}...\left(1\right)$

$x^2+y^2=25...\left(2\right)$

Put (1) in (2)

$x^2+(\frac{10}{x}{)}^2=25$

$x^4+100=25x^2$

$x^4-25x^2+100=0$

$x^4-20x^2-5x^2+100=0$

$x^2(x^2-20)-5(x^2-20)=0$

$\Rightarrow (x^2-20)(x^2-5)=0$

$\Rightarrow x^2=20,x^2=5$

$\Rightarrow x=\pm \sqrt{20},x=\pm \sqrt{5}$

$\therefore$ there are 4 real solutions.