Question

solve the following equation by quadratic formula:
$\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}\,\,;\,x \ne \, - 1, - 2, - 4$

Answer 1

Hint: This is a simple question of the quadratic equation. We just have to find the roots of our quadratic equations. But first, we have to make a quadratic equation. Then we can solve our problem either by completing the square method or dharacharya formula. One more important thing to remember is that whenever we take the root of any number, it comes with positive and negative signs.
Formula used: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where $D = \sqrt {{b^2} - 4ac} $

Complete step by step answer:
In the given question,
$\dfrac{1}{{x + 1}} + \dfrac{2}{{x + 2}} = \dfrac{4}{{x + 4}}$
Taking L.C.M,
$\dfrac{{\left( {x + 2} \right) + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{x + 4}}$
$\dfrac{{\left( {x + 2} \right) + \left( {2x + 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{\left( {x + 4} \right)}}$
On adding,
$\dfrac{{\left( {3x + 4} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \dfrac{4}{{\left( {x + 4} \right)}}$
On cross-multiplication, we get
$\left( {3x + 4} \right)\left( {x + 4} \right) = 4\left( {x + 1} \right)\left( {x + 2} \right)$
On multiplying both sides,
$3{x^2} + 12x + 4x + 16 = 4\left( {{x^2} + 2x + x + 2} \right)$
$3{x^2} + 12x + 4x + 16 = 4{x^2} + 8x + 4x + 2$
Cancel $4x$ from both sides
$3{x^2} + 12x + 16 = 4{x^2} + 8x + 2$
$4{x^2} - 3{x^2} + 8x - 12x + 2 - 16 = 0$
${x^2} - 4x - 14 = 0$
Now,
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where x is the root of the quadratic equation $a{x^2} + bx + c = 0$
Also, $D = \sqrt {{b^2} - 4ac} $
On comparing with the given equation,
$a = 1$
$b = - 4$
$c = - 14$
Now, calculate D
$D = \sqrt {{b^2} - 4ac} $
$D = {\left( { - 4} \right)^2} - 4 \times 1 \times \left( { - 14} \right)$
$D = 16 + 56$
$D = 72$
Now, Substituting the values of D, b and a in the above formula.
$x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {72} }}{{2\left( 1 \right)}}$
$x = \dfrac{{4 \pm 6\sqrt 2 }}{2}$
$x = \dfrac{4}{2} \pm \dfrac{{6\sqrt 2 }}{2}$
$x = 2 \pm 3\sqrt 2 $
Hence, the values of x are $2 + 3\sqrt 2 $ and $2 - 3\sqrt 2 $ .

Note:
Apart from the dharacharya formula we can also solve this question by completing the square method. Also, be careful while solving a question whenever restrictions are given. In this question, we can’t take the values of x as $ - 1,\, - 2$ and $ - 4$ as they are making the denominator zero and we can’t take these values in our answer if they are coming out to be.