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3 years ago

12

Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j^2 + 3j – 28 = 0

1 answer:

tresset_1 [31]3 years ago

8 0

$\bf \begin{array}{lcccll}
& -4j^2& +3j& -28\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad \qquad discriminant\implies b^2-4ac
\\\\\\
(3)^2-4(-4)(-28)=
\begin{cases}
0&\textit{one solution}\\
positive&\textit{two solutions}\\
negative&\textit{no solution}
\end{cases}$

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