Question

Need help on #30 and 31 thanks!!

Answer 1

30)

$\bf \begin{array}{llll} y=&{{ a}}x^2&{{ +b}}x&{{ +c}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \\\\\\ discriminant\implies b^2-4ac= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases}$

31)


a simple case for that would just be, using an equation with an imaginary value, let's do so

$\bf \sqrt{-5}=x\implies \sqrt{-1\cdot 5}=x\implies \sqrt{-1}\sqrt{5}=x\implies i\sqrt{5}=x\\\\ -------------------------------\\\\ \textit{so, we'll use that imaginary value then}\\\\ \sqrt{-5}=x\implies -5=x^2\implies 0=x^2+5\implies \boxed{y=x^2+5}$

when you get a "solution" or zero with an "i" or an imaginary value, is just a way to say, there's really no solution, the function never touches the x-axis