Question

What value is a discontinuity of x squared plus 2 x plus 3, all over x squared minus x minus 12 ?

Answer 1

Answer:

$x=-3, x=4$

Step-by-step explanation:

we have

$\frac{x^{2}+2x+3}{x^{2}-x-12}$

we know that

The denominator can not be zero

Find the roots of the quadratic equation of denominator

$x^{2}-x-12=0$

using a graphing tool to resolve the quadratic equation

see the attached figure

The solutions are

$x=-3, x=4$

so

$x^{2}-x-12=(x+3)(x-4)$

therefore

we have

$\frac{x^{2}+2x+3}{(x+3)(x-4)}$

The values $x=-3, x=4$ are discontinuity of the original function

Answer 2

X² + 2x +3

If the discriminent Δ = b²-4.a.c <0 , that means the roots (or the zero value) are nit REAL Number (they are imaginary or complex) 

Δ = 2² - 4(1)3) = 4 - 12 = -8

since Δ = - 8 < 0 there are no real roots