If $x+y=3$ and $\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}$, then the value of $\left(x^2+y^2\right)$ is :
Answer & explanation
Correct answer: option 1
$x+y=3$
$\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}$
\(\frac{y + x}{xy}\) = -\(\frac{3}{10}\)
\(\frac{3}{xy}\) = -\(\frac{3}{10}\)
xy = -10
( a + b )2 = a2 + b2 + 2ab
$\left(x^2+y^2\right)$ = ( 3 )2 - 2 × -10
$\left(x^2+y^2\right)$ = 29