## Precalculus (10th Edition)

a) See graph; b) $\{(-1.811,9.835),(1.811,9.835)\}$ c) $\{(-1.811,9.835),(1.811,9.835)\}$
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: $\{(-1.811,9.835),(1.811,9.835)\}$