Answer
a) See graph;
b) $\{(-1.811,9.835),(1.811,9.835)\}$
c) $\{(-1.811,9.835),(1.811,9.835)\}$
Work Step by Step
a) Graph the two curves:
$x^2+y^2=100$
$y=3x^2$
b) We have to solve the system:
$\begin{cases}
x^2+y^2=100\\
y=3x^2
\end{cases}$
Use substitution:
$x^2=\dfrac{y}{3}$
$\dfrac{y}{3}+y^2=100$
$y+3y^2=300$
$3y^2+y-300=0$
$y=\dfrac{-1\pm\sqrt{1^2-4(3)(-300)}}{2(3)}\approx\dfrac{-1\pm 60.0083}{6}$
$y_1=\dfrac{-1-60.0083}{6}\approx -10.168$
$y_2=\dfrac{-1+60.0083}{6}\approx 9.835$
As $y=3x^2\geq 0$, only one solution fits:
$y\approx 9.835$
Determine $x$:
$x^2=\dfrac{9.835}{3}$
$x=\pm\sqrt{\dfrac{9.835}{3}}=\approx \pm1.811$
The solution set is:
$\{(-1.811,9.835),(1.811,9.835)\}$
c) From the graph we find the intersection points:
$\{(-1.811,9.835),(1.811,9.835)\}$
