How do you solve the following Quadratic Inequality #x^2+2x-15

Question

How do you solve the following Quadratic Inequality #x^2+2x-15<0#?

Cora Alexander · Accepted Answer

The solution is
#-5 < x < 3# There are more than one ways to solve this inequality. Solution #A)# Since #x^2+2x-15 = (x+5)(x-3)#, we can suggest the following reasoning. The product of two real numbers can be negative if one of them is positive and another is negative. Therefore, we have two solutions: #A_1)# #x+5 > 0# AND #x-3 < 0# #A_2)# #x+5 < 0# AND #x-3 > 0# The case #A_1# defines #x# as #x > -5# AND #x < 3#, which defines an interval for #x#: #-5 < x < 3# The case #A_2# defines #x# as #x < -5# AND #x > 3#, which is impossible. So, the solution is #-5 < x < 3# Solution #B)# As we know, the graph of the quadratic polynomial on the left is parabola. Since the coefficient at #x^2# is positive, this parabola directs its endpoints upward. Therefore, the only way it can be negative is in-between its roots, where it's equal to zero. In other words, the solutions to a inequality is the area between solutions to equality #x^2+2x-15=0# Obvious solutions are #x=3# and #x=-5#. So, the solutions are #-5 < x < 2#

Madelyn Anderson · Answer

undefined To solve the quadratic inequality $x^2 + 2x - 15 < 0$: 1. Find the critical points by setting the quadratic expression equal to zero and solving for $x$. $x^2 + 2x - 15 = 0$ Factor the quadratic expression or use the quadratic formula to find the roots. $x^2 + 5x - 3x - 15 = 0$ $x(x + 5) - 3(x + 5) = 0$ $(x - 3)(x + 5) = 0$ $x = 3$ or $x = -5$ 2. Plot these critical points on the number line. $\begin{array}{cccccccc} - & -5 & - & 3 & + \ \end{array}$ 3. Choose test points within each interval formed by the critical points and evaluate the quadratic expression $x^2 + 2x - 15$ at these points. Test point in $(- \infty, -5)$: $x = -6$, $x^2 + 2x - 15 = (-6)^2 + 2(-6) - 15 = 36 - 12 - 15 = 9 > 0$ Test point in $(-5, 3)$: $x = 0$, $x^2 + 2x - 15 = (0)^2 + 2(0) - 15 = -15 < 0$ Test point in $(3, +\infty)$: $x = 4$, $x^2 + 2x - 15 = (4)^2 + 2(4) - 15 = 16 + 8 - 15 = 9 > 0$ 4. Determine the sign of the quadratic expression within each interval. In $(- \infty, -5)$, the expression is positive ($> 0$). In $(-5, 3)$, the expression is negative ($< 0$). In $(3, +\infty)$, the expression is positive ($> 0$). 5. Determine the solution to the inequality based on the signs determined in step 4. The solution to $x^2 + 2x - 15 < 0$ is $x$ belonging to the interval $(-5, 3)$.

How do you solve the following Quadratic Inequality #x^2+2x-15&lt;0#?

How do you solve the following Quadratic Inequality #x^2+2x-15<0#?