Question

Solve for x: \[\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}\]

Answer 1

Hint: Simplifying such equations will give us higher degree equations.
If the equation is a quadratic equation we can either use a splitting middle term method or can use a discriminant method to solve the equation.
Discriminant method: For a quadratic equation ${ax^2}$+bx+c=0, the value of x will be: \[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{2}\]

Complete step-by-step answer:
Given equation
\[\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}\]
Taking L.CM and simplifying the equation we get;
\[ \Rightarrow \dfrac{{x + 2 + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}\]
\[ \Rightarrow \dfrac{{x + 2 + 2x + 2}}{{{x^2} + 2x + x + 2}} = \dfrac{4}{{x + 4}}\]
\[ \Rightarrow \dfrac{{3x + 4}}{{{x^2} + 3x + 2}} = \dfrac{4}{{x + 4}}\]
Cross- multiplying both sides we get;
\[ \Rightarrow \left( {3x + 4} \right)\left( {x + 4} \right) = 4{x^2} + 12x + 8\]
\[ \Rightarrow 3{x^2} + 12x + 4x + 16 = 4{x^2} + 12x + 8\]
\[ \Rightarrow 3{x^2} + 16x + 16 = 4{x^2} + 12x + 8\]
\[ \Rightarrow {x^2} - 4x - 8 + 0\]
Above equation is a quadratic equation.
Using discriminant method;
\[ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 32} }}{2}\]
\[ \Rightarrow x = \dfrac{{4 \pm 4\sqrt 3 }}{2}\]
\[ \Rightarrow x = 2 \pm 2\sqrt 3 \]
Required value for (x) =\[2 \pm 2\sqrt 3 \]

Note: A quadratic equation has degree 2 so the number of roots are also 2.
Depending on the value of (d), the number of roots can be real and distinct, real and equal or complex.
When D > 0, the roots are real and distinct.
When D =0, the roots are real and equal.
When D < 0, the roots are complex.