A) Given:
[tex]\begin{gathered} x^2+y^2=17\ldots\ldots\ldots\text{.}(1) \\ y=-\frac{1}{2}x\ldots\ldots\ldots\ldots(2) \end{gathered}[/tex]
To find: The number of real solutions
Explanation:
Substitute equation (2) in (1), we get
[tex]\begin{gathered} x^2+(-\frac{1}{2}x)^2=17 \\ x^2+\frac{x^2}{4}=17 \\ \frac{5x^2}{4}=17^{} \\ x^2=\frac{68}{5} \\ x^2-\frac{68}{5}=0\ldots\ldots.(3) \end{gathered}[/tex]
Here,
[tex]a=1,b=0,\text{ and c=-}\frac{68}{5}[/tex]
So, the discriminant is,
[tex]\begin{gathered} \Delta=b^2-4ac \\ =0-4(1)(-\frac{68}{5}) \\ =54.4 \\ >0 \end{gathered}[/tex]
Since the discriminant is greater than zero.
Hence, it has two real solutions
Final answer:
System A has two real solutions.
