How do you solve and write the following in interval notation: #x^2+4x-120#?

Question

How do you solve and write the following in interval notation: #x^2+4x-12>0#?

Eva Adamson · Accepted Answer

#x in RR: (-oo,-6)uu(+2,+oo)#

#x^2+4x-12>0#

To solve this inequality and write in interval notation we must find the values of #x in RR# for which #x^2+4x-12>0#

Let's first consider the critical points for which #x^2+4x-12=0#

#x^2+4x-12 = (x+6)(x-2)# which #= 0# for #x=-6 or x=2#

We know #x^2+4x-12# is a parabola with a minimum value (since the coefficient of #x^2 >0#

Hence, #x^2+4x-12>0# for all real #x# from #-oo# up to #-6# exclusive and from #+2# exclusive up to #+oo#

This can be written as: #x in RR: (-oo,-6)uu(+2,+oo)#

We can see this from the graph of #x^2+4x-12# below:

graph{x^2+4x-12 [-26.85, 24.5, -17.23, 8.45]}

Avery Alexander · Answer

undefined To solve the inequality $x^2 + 4x - 12 > 0$, first find the critical points by setting the expression equal to zero and solving for $x$. Then, use these critical points to create intervals and test points within each interval to determine where the inequality holds true. Finally, write the solution in interval notation.

1. **Find critical points**: Set $x^2 + 4x - 12 = 0$ and solve for $x$.
   
   $x^2 + 4x - 12 = 0$ factors as $(x + 6)(x - 2) = 0$.
   
   The critical points are $x = -6$ and $x = 2$.

2. **Create intervals**: Use the critical points to divide the number line into intervals: $(- \infty, -6)$, $(-6, 2)$, and $(2, +\infty)$.

3. **Test each interval**: Choose a test point within each interval and evaluate $x^2 + 4x - 12$ to determine if it's greater than zero.

- Test $x = -7$ in the interval $(- \infty, -6)$: $(-7)^2 + 4(-7) - 12 = 49 - 28 - 12 = 9 > 0$.
   - Test $x = 0$ in the interval $(-6, 2)$: $0^2 + 4(0) - 12 = -12 < 0$.
   - Test $x = 3$ in the interval $(2, +\infty)$: $3^2 + 4(3) - 12 = 9 + 12 - 12 = 9 > 0$.

4. **Write solution in interval notation**: The inequality holds true when $x$ is in the intervals where $x^2 + 4x - 12 > 0$, which are $(- \infty, -6)$ and $(2, +\infty)$.

So, in interval notation, the solution is: $(- \infty, -6) \cup (2, +\infty)$.

How do you solve and write the following in interval notation: #x^2+4x-12&gt;0#?

How do you solve and write the following in interval notation: #x^2+4x-12>0#?