Question

Solve the following quadratic equation by factorization method:
${\left( {{\text{a + b}}} \right)^2}{{\text{x}}^2} - 4{\text{abx - }}{\left( {{\text{a - b}}} \right)^2} = 0$

Answer 1

Hint: In order to solve the following quadratic equation, we have to divide the entire equation by a term of the equation. Then we express the equation as one term is equal to the other. Later we express both sides of the equation as square roots and solve for the value of x. This gives us the factors of the given equation.

Complete step by step answer:
Given data,
To factorize${\left( {{\text{a + b}}} \right)^2}{{\text{x}}^2} - 4{\text{abx - }}{\left( {{\text{a - b}}} \right)^2} = 0$.

We factorize the given equation by the factorization method as follows:
${\left( {{\text{a + b}}} \right)^2}{{\text{x}}^2} - 4{\text{abx - }}{\left( {{\text{a - b}}} \right)^2} = 0$
Let us divide the entire equation by the term${\left( {{\text{a + b}}} \right)^2}$, we get
\[
   \Rightarrow {{\text{x}}^2} - \dfrac{{4{\text{abx}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}{\text{ - }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}} = 0 \\
   \Rightarrow {{\text{x}}^2} - \dfrac{{4{\text{ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}{\text{x = }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}} \\
\]
Now let us add the term$\dfrac{{{\text{4}}{{\text{a}}^2}{{\text{b}}^2}}}{{{{\left( {{\text{a + b}}} \right)}^4}}}$on both sides, we get
\[ \Rightarrow {{\text{x}}^2} - \dfrac{{4{\text{ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}{\text{x + }}\dfrac{{{\text{4}}{{\text{a}}^2}{{\text{b}}^2}}}{{{{\left( {{\text{a + b}}} \right)}^4}}}{\text{ = }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}} + \dfrac{{{\text{4}}{{\text{a}}^2}{{\text{b}}^2}}}{{{{\left( {{\text{a + b}}} \right)}^4}}}\]
The LHS of the equation looks in the form of expansion of the term${\left( {{\text{a - b}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - 2ab}}$.
Where x = a and b = \[\dfrac{{{\text{2ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}\]

Comparing this to the LHS, we have, a = x, ${\text{b = }}\dfrac{{{\text{2ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}$
\[
   \Rightarrow {\left( {{\text{x}} - \dfrac{{{\text{2ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}} \right)^2}{\text{ = }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}{{\left( {{\text{a + b}}} \right)}^2}{\text{ + 4}}{{\text{a}}^2}{{\text{b}}^2}}}{{{{\left( {{\text{a + b}}} \right)}^4}}} \\
   \Rightarrow {\left( {{\text{x}} - \dfrac{{{\text{2ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}} \right)^2}{\text{ = }}{\left( {\dfrac{{\left( {{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}} \right)}}{{{{\left( {{\text{a + b}}} \right)}^2}}}} \right)^2} \\
\]
Now let us take the square root on both sides, we get
\[
   \Rightarrow {\text{x}} - \dfrac{{{\text{2ab}}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}{\text{ = }} \pm \dfrac{{\left( {{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}} \right)}}{{{{\left( {{\text{a + b}}} \right)}^2}}} \\
   \Rightarrow {\text{x = }}\dfrac{{{\text{2ab }} \pm {\text{ }}\left( {{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}} \right)}}{{{{\left( {{\text{a + b}}} \right)}^2}}} \\
\]
We know the expansion of the term${\left( {{\text{a - b}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - 2ab}}$and${\left( {{\text{a + b}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + 2ab}}$. Using them in the above we get,
\[
   \Rightarrow {\text{x = }}\dfrac{{{{\left( {{\text{a + b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}},{\text{ - }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}} \\
   \Rightarrow {\text{x = 1}},{\text{ - }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}} \\
\]
Therefore the factors of the equation ${\left( {{\text{a + b}}} \right)^2}{{\text{x}}^2} - 4{\text{abx - }}{\left( {{\text{a - b}}} \right)^2} = 0$ are (x – 1) and $\left( {{\text{x + }}\dfrac{{{{\left( {{\text{a - b}}} \right)}^2}}}{{{{\left( {{\text{a + b}}} \right)}^2}}}} \right)$.

Note: In order to solve this type of questions the key is to know the expansion of the terms of the form${\left( {{\text{a - b}}} \right)^2}{\text{ and }}{\left( {{\text{a + b}}} \right)^2}$.
To perform the factorization method the RHS of the quadratic equation should be equal to zero. We start off this method by removing all the common terms of the equation, then we perform different things like arithmetic operations on the terms, using the formulae of expansions to further simplify it.
The square root of a variable can take both the positive and negative values of the variable.