Question

Find the roots of the equation \[{{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0\] by the method of completing the square.

Answer 1

Hint: In the given question, we have been asked to solve the given quadratic equation by the method of completing the square. We will start by dividing the whole quadratic equation by \[{{a}^{2}}\] . Then we will rewritten the equation in a form it will be easy to solve. Later we will add and subtract \[{{\left( \dfrac{3b}{2a} \right)}^{2}}\] . Then we will using the identity of numbers i.e. \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] and then further solve the equation for the roots of the equation i.e. values of ‘x’.
Identity used: \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]

Complete step by step solution:
We have given that,
 \[{{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0\]
Dividing both the sides of the given equation by \[{{a}^{2}}\] , we will get
 \[\Rightarrow \dfrac{{{a}^{2}}{{x}^{2}}}{{{a}^{2}}}-\dfrac{3abx}{{{a}^{2}}}+\dfrac{2{{b}^{2}}}{{{a}^{2}}}=\dfrac{0}{{{a}^{2}}}\]
Cancelling out the common terms from the above equation, we will get
 \[\Rightarrow {{x}^{2}}-\dfrac{3bx}{a}+\dfrac{2{{b}^{2}}}{{{a}^{2}}}=0\]
Rewrite the given equation as,
 \[\Rightarrow {{x}^{2}}-2\left( \dfrac{3b}{2a} \right)\left( x \right)+\dfrac{2{{b}^{2}}}{{{a}^{2}}}=0\]
Subtracting \[\dfrac{2{{b}^{2}}}{{{a}^{2}}}\] from both the sides of the equation, we will get
 \[\Rightarrow {{x}^{2}}-2\left( \dfrac{3b}{2a} \right)\left( x \right)=-\dfrac{2{{b}^{2}}}{{{a}^{2}}}\]
Adding and subtracting \[{{\left( \dfrac{3b}{2a} \right)}^{2}}\] on the LHS of the above equation,
 \[\Rightarrow {{x}^{2}}-2\left( \dfrac{3b}{2a} \right)\left( x \right)+{{\left( \dfrac{3b}{2a} \right)}^{2}}-{{\left( \dfrac{3b}{2a} \right)}^{2}}=-\dfrac{2{{b}^{2}}}{{{a}^{2}}}\]
Taking \[-{{\left( \dfrac{3b}{2a} \right)}^{2}}\] on the RHS of the equation, we will get
(Remember that on transposing the sign will be changed from negative to positive.)
 \[\Rightarrow {{x}^{2}}-2\left( \dfrac{3b}{2a} \right)\left( x \right)+{{\left( \dfrac{3b}{2a} \right)}^{2}}=-\dfrac{2{{b}^{2}}}{{{a}^{2}}}+{{\left( \dfrac{3b}{2a} \right)}^{2}}\]
Using the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] in the above equation, we will get
 \[\Rightarrow {{\left( x-\left( \dfrac{3b}{2a} \right) \right)}^{2}}=-\dfrac{2{{b}^{2}}}{{{a}^{2}}}+{{\left( \dfrac{3b}{2a} \right)}^{2}}\]
Simplifying the numbers in the above equation,
 \[\Rightarrow {{\left( x-\left( \dfrac{3b}{2a} \right) \right)}^{2}}=-\dfrac{2{{b}^{2}}}{{{a}^{2}}}+\dfrac{9{{b}^{2}}}{4{{a}^{2}}}\]
Taking the LCM of the denominators on the RHS of the equation and simplifying,
 \[\Rightarrow {{\left( x-\left( \dfrac{3b}{2a} \right) \right)}^{2}}=\dfrac{-8{{b}^{2}}+9{{b}^{2}}}{4{{a}^{2}}}\]
 \[\Rightarrow {{\left( x-\left( \dfrac{3b}{2a} \right) \right)}^{2}}=\dfrac{{{b}^{2}}}{4{{a}^{2}}}\]
Transposing the power of 2 to the RHS of the above equation, we will get
 \[\Rightarrow \left( x-\left( \dfrac{3b}{2a} \right) \right)=\sqrt{\dfrac{{{b}^{2}}}{4{{a}^{2}}}}\]
Simplifying the above, we will get
 \[\Rightarrow x-\dfrac{3b}{2a}=\pm \dfrac{b}{2a}\]
Adding \[\dfrac{3b}{2a}\] to both the sides of the equation,
 \[\Rightarrow x=\dfrac{3b}{2a}\pm \dfrac{b}{2a}\]
Thus, we will get two values of the ‘x’,
 \[\Rightarrow x=\dfrac{3b}{2a}+\dfrac{b}{2a},\ \dfrac{3b}{2a}-\dfrac{b}{2a}\]
Solving the values, we will get
 \[\Rightarrow x=\dfrac{4b}{2a},\ \dfrac{2b}{2a}\]
Cancelling out the common terms from both the values, we will get
 \[\Rightarrow x=\dfrac{2b}{a},\ \dfrac{b}{a}\]
Therefore,
The solution set of the equation \[{{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0\] is \[x\in \left\{ \dfrac{b}{a},\ \dfrac{2b}{a} \right\}\] .
So, the correct answer is “ \[x\in \left\{ \dfrac{b}{a},\ \dfrac{2b}{a} \right\}\] ”.

Note: In the above question, it is important to note that we have used a fact that is if \[{{x}^{2}}=a\] then \[x=\pm \sqrt{a}\] . Therefore this implies that we get two values of ‘x’ that is one value is negative and the other value is positive. It is important to add the sign \['\pm '\] , so that we do not miss any of the solutions. It is a very common mistake while solving the quadratic equation, so we need to keep this in mind.