Question

Solve this equation $\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}$, where $x + 1 \ne 0$, $x + 2 \ne 0$ and $x + 4 \ne 0$ using quadratic formula.

Answer 1

Hint: Here, in the given question, it is mentioned to solve the equation by using the quadratic formula. The quadratic formula is a formula that provides the solution(s) to a quadratic equation. Quadratic equations are second-degree algebraic expressions and are of the form $a{x^2} + bx + c = 0$. To solve the given question, we will first take LCM on the left-hand side of the equation given in the question. After simplifying the equation, we will get a quadratic equation in $x$. After this, we will apply the quadratic formula to solve the equation.
Formulae used: Quadratic formula: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
We have,
$\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}$
We will take LCM on the left-hand side of the above equation. Thus, we will get the following equation.
$ \Rightarrow \dfrac{{1\left( {x + 2} \right) + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}$
On multiplication of terms in brackets, we get
$ \Rightarrow \dfrac{{x + 2 + 2x + 2}}{{{x^2} + 2x + x + 2}} = \dfrac{4}{{x + 4}}$
$ \Rightarrow \dfrac{{3x + 4}}{{{x^2} + 3x + 2}} = \dfrac{4}{{x + 4}}$
On cross-multiplication, we get
$ \Rightarrow \left( {3x + 4} \right)\left( {x + 4} \right) = 4\left( {{x^2} + 3x + 2} \right)$
$ \Rightarrow 3{x^2} + 12x + 4x + 16 = 4{x^2} + 12x + 8$
Shift all the terms on one side
$ \Rightarrow 4{x^2} - 3{x^2} + 12x - 12x - 4x + 8 - 16 = 0$
After addition and subtraction of like terms, we get
$ \Rightarrow {x^2} - 4x - 8 = 0$
Now, we have our equation in the form of quadratic equation, that is $a{x^2} + bx + c = 0$, $a \ne 0$
For the quadratic equation ${x^2} - 4x - 8 = 0$ we have, $a = 1$, $b = - 4$ and $c = - 8$.
Now, we will substitute these values in the quadratic formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( { - 8} \right)} }}{{2\left( 1 \right)}}$
On simplification, we get
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 4\left( { - 8} \right)} }}{2}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 32} }}{2}$
On solving the terms present inside the square root, we get
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {48} }}{2}$
On solving square root of $48$, we get $\sqrt {48} = \sqrt {2 \times 2 \times 2 \times 2 \times 3} = 4\sqrt 3 $.
$ \Rightarrow x = \dfrac{{4 \pm 4\sqrt 3 }}{2}$
Take $2$ as a common factor from numerator and denominator
$ \Rightarrow x = \dfrac{{2\left( {2 \pm \sqrt 3 } \right)}}{2}$
On cancelling $2$from numerator and denominator, we get
$ \Rightarrow x = 2 \pm 2\sqrt 3 $
We can write roots of given equation as $\alpha = 2 + 2\sqrt 3 $ and $\beta = 2 - 2\sqrt 3 $.

Note: A quadratic equation of the form $a{x^2} + bx + c = 0$ where $a \ne 0$ can be solved by many methods. Such as, completing the square method, factorization method, by using the quadratic formula, etc. Here, in the given question a particular method is mentioned, so we solved it by that method only. Doesn’t matter with which method we solve any quadratic equation because the roots of the quadratic equation will remain the same.