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Find the break even points of $\frac{x^2+x-2}{x^2+5x+6}$

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Solution

Final Answer

$x=1$
Got another answer? Verify it here!

Step-by-step Solution

Problem to solve:

$\frac{x^2+x-2}{x^2+5x+6}=0$

Specify the solving method

1

Find the break even points of the polynomial $\frac{x^2+x-2}{x^2+5x+6}$ by putting it in the form of an equation and then set it equal to zero

$\frac{x^2+x-2}{x^2+5x+6}=0$
2

Factor the trinomial $x^2+5x+6$ finding two numbers that multiply to form $6$ and added form $5$

$\begin{matrix}\left(2\right)\left(3\right)=6\\ \left(2\right)+\left(3\right)=5\end{matrix}$
3

Thus

$\frac{x^2+x-2}{\left(x+2\right)\left(x+3\right)}=0$
4

Factor the trinomial $x^2+x-2$ finding two numbers that multiply to form $-2$ and added form $1$

$\begin{matrix}\left(-1\right)\left(2\right)=-2\\ \left(-1\right)+\left(2\right)=1\end{matrix}$
5

Thus

$\frac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}=0$
6

Simplifying

$\frac{x-1}{x+3}=0$
7

Multiply both sides of the equation by $x+3$

$x-1=0$
8

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-1$ from both sides of the equation

$x-1-1\cdot -1=0-1\cdot -1$
9

Multiply $-1$ times $-1$

$x-1+1=0-1\cdot -1$
10

Multiply $-1$ times $-1$

$x-1+1=0+1$

Verify that the solutions obtained are valid in the initial equation

11

The valid solutions to the equation are the ones that, when replaced in the original equation, don't make any denominator equal to $0$, since division by zero is not allowed

$x=1$

Final Answer

$x=1$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve for xFind the rootsSolve by factoringSolve by completing the squareSolve by quadratic formulaFind the discriminant

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7
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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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