Question

Where is the removable discontinuity of f(x)=(x+5)/(2x^2+3x-10) located
a.x=-5
b.x = –2
c.x = 2
d.x = 5

Answer 1

please can you check your formula of f(x)?

Answer 2

is the denominator, like this:
\[2{x^2} + 3x - 10\]

Answer 3

yes

Answer 4

the solution of this equation:
\[2{x^2} + 3x - 10 = 0\]
are:
\[\begin{gathered}
x = \frac{{ - 3 \pm \sqrt {{3^2} - 4 \times 2 \times \left( { - 10} \right)} }}{{2 \times 2}} = \hfill \\
\hfill \\
= \frac{{ - 3 \pm \sqrt {9 + 80} }}{4} = \frac{{ - 3 \pm \sqrt {89} }}{4} \hfill \\
\end{gathered} \]
am I right?

Answer 5

I have applied the standard formula:
\[x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\]

Answer 6

I think that the denominator can be this:
\[{x^2} + 3x - 10\]
is it possible?