find all values of k such that x^2-kx+9 = 0 has two imaginary roots
do u know discriminant of this equaation
6
but thats not right apparently
formula for discriminant \(\large \color{black}{\begin{align} if\quad ax^2+bx+c=0\hspace{.33em}\\~\\ then\quad D=b^2-4ac\hspace{.33em}\\~\\ \end{align}}\)
can u find \(D\) now?
-k^2-36
some fault there , note that \(\large b=-k\) here
yeah and i said (-k)^2
\(-k^2\neq (-k)^2\)
\((-k)^2=k^2\)
ok
ok the discriminant property says that the equation has imaginary roots if \(\large \color{black}{\begin{align} D<0\hspace{.33em}\\~\\ \end{align}}\)
yes
so \(\large \color{black}{\begin{align} k^2-36<0\hspace{.33em}\\~\\ \end{align}}\) can u solve this
k< 6
that's not complete, use the property that \(\large \color{black}{\begin{align} a^2-b^2=(a-b)(a+b)\hspace{.33em}\\~\\ \end{align}}\)
can you tell me what you got
i got \(\large \color{black}{\begin{align} k^2-36<0\hspace{.33em}\\~\\ k^2-6^2<0\hspace{.33em}\\~\\ \end{align}}\)
isnt it k> -6
now we have \(\large \color{black}{\begin{align} k^2-6^2<0\hspace{.33em}\\~\\ (k-6)(k+6)<0\hspace{.33em}\\~\\ \end{align}}\)
so we have \(\large \color{black}{\begin{align} k<6\hspace{.33em}\\~\\ and \hspace{.33em}\\~\\ k>-6\hspace{.33em}\\~\\ \end{align}}\)
so \(\large \color{black}{\begin{align} -6<k<6\hspace{.33em}\\~\\ \end{align}}\)
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