Question

Solve the equation for $x$:
$\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}\;\;\;\;,x \ne - 1, - 2, - 4$

Answer 1

Here we will use the LCM method by multiplying and dividing the two polynomial fractions in the LHS side using common factors and ensure that denominators are the same.
We will find the roots of $x$ by using the quadratic equation formula i.e. for any given quadratic equation $a{x^2} + bx + c = 0$, the roots of $x$ is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where $a$ is the coefficient of ${x^2}$,$b$ is the coefficient of $x$ and $c$ is the constant term.

Complete step-by-step answer:
Step 1: For the given equation, $\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}$, by multiplying and dividing the two polynomial fractions in the LHS side using common factors we ensure that denominators are the same.
$\dfrac{{1 \cdot \left( {x + 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} + \dfrac{{2 \cdot \left( {x + 1} \right)}}{{\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}$…. (1)
Step 2: Now, in the above equation (1) by multiplying the factors with $1$ and $2$ in LHS side and then adding the two fractions we get:
$\Rightarrow \dfrac{{x + 2 + 2x + 2}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}$…. (2)
Step 3: Now, by adding the coefficients of $x$ and the constants in LHS in the equation (2) we get:
$\Rightarrow \dfrac{{3x + 4}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}$…. (3)
Step 4: For the equation $\dfrac{{3x + 4}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}$, by multiplying the denominator of the RHS side which is $x + 4$with the numerator of the LHS side which is $3x + 4$. Similarly, by multiplying the denominator of the LHS side which is $\left( {x + 1} \right)\left( {x + 2} \right)$with the numerator of the RHS side which is $4$.
$\Rightarrow \left( {3x + 4} \right)\left( {x + 4} \right) = 4\left( {x + 1} \right)\left( {x + 2} \right)$ …. (4)
Step 4: Now, simplifying the above equation (4):
First of all, opening the brackets and multiplying them to make an equation:

$$\left( {3x + 4} \right)\left( {x + 4} \right) = 4\left( {x + 1} \right)\left( {x + 2} \right) \\\ \Rightarrow 3{x^2} + 12x + 4x + 16 = 4\left( {{x^2} + 2x + 1x + 2} \right) \\\$$

In the RHS side, by multiplying each factor by 4:
$\Rightarrow 3{x^2} + 16x + 16 = 4{x^2} + 8x + 4x + 8$…. (5)
Step 6: For the equation (5), we will get the variable $x$ in LHS to form a quadratic equation by taking RHS to the LHS side:
$\Rightarrow - {x^2} + 4x + 8 = 0$… (6)
Step 7: Now, by solving the above equation (6), using a quadratic formula that for $a{x^2} + bx + c = 0$, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
$\Rightarrow x = \dfrac{{ - 4 \pm \sqrt {{4^2} - \left( {4 \times \left( { - 1} \right) \times 8} \right)} }}{{2 \times \left( { - 1} \right)}}$ , where $b = 4,a = - 1,c = 8$. ….. (7)
Step 8: As we know, ${4^2} = 16$, $2 \times \left( { - 1} \right) = - 2$ and $4 \times \left( { - 1} \right) \times 8 = - 32$, substituting these values in the above equation (7):
$x = \dfrac{{ - 4 \pm \sqrt {{4^2} - \left( {4 \times \left( { - 1} \right) \times 8} \right)} }}{{2 \times \left( { - 1} \right)}}$:
By putting the value as given:
$\Rightarrow \dfrac{{ - 4 \pm \sqrt {16 - \left( { - 32} \right)} }}{{ - 2}} = \dfrac{{ - 4 \pm \sqrt {48} }}{{ - 2}}$ …..…. (8)
Step 9: Now, we know that the factors of $48 = 4 \times 4 \times 3$, so $\sqrt {48} = 4\sqrt 3$ . So, using this result in the above term (8):

$$x = \dfrac{{ - 4 \pm \sqrt {48} }}{{ - 2}} \\\ = \dfrac{{ - 4 \pm 4\sqrt 3 }}{{ - 2}} \\\$$

Step 10: Now, Dividing $- 2$ in the denominator with the numerator $- 4 \pm 4\sqrt 3$ we will get the final result $x$ from the equation $x = \dfrac{{ - 4 \pm 4\sqrt 3 }}{{ - 2}}$.
$\Rightarrow x = 2 \pm 2\sqrt 3$
Step 11: Since $x = 2 \pm 2\sqrt 3$ so we have two values of $x$, $x = 2 + 2\sqrt 3$ and $x = 2 - 2\sqrt 3$. Also, we notice that in the result thus obtained $x \ne - 1, - 2, - 4$ satisfying the conditions given in the question.

So, there are two possible values of $x$ are $x = 2 + 2\sqrt 3$ and $x = 2 - 2\sqrt 3$

Note: Students used to make mistakes in calculating the value of $x$. You should also remember that since there are three $x$ terms in the equation $\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}\;\;\;\;,x \ne - 1, - 2, - 4$ , so do not take LCM of all denominators in one go. This will complicate the equation.