Question

Consider the following equation and values of x: ${{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0,\,\,\,\,x=\dfrac{a}{b},\,\,\,\,x=\dfrac{b}{a}$ . Find the right solution to the equation.

Answer 1

Hint: We are given a equation as \[{{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0\] we are asked to find solution, we will learn about type of equation then we will learn how meaningful solution are possible for an equation, we will then use middle term split method in which we will split middle term of our equation to get the equation into factor then lastly we will use zero product rule to get the solution.

Complete step by step solution:
We are given ${{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0$ we can see that the hare a, b is constant at n is variable, whose highest power is $2$. So, a $2$ degree equation means a quadratic equation. We know the quadratic equation has at most $2$ solution.
Now we will split our equation into a middle term split method. In this method we find those pair of numbers whose difference is same as the number b (middle term) of ${{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0$ and their product is same as a and c. Now in an equation ${{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0$ we have
$a={{a}^{2}},\,b=-3ab,\,c=2{{b}^{2}}$
So, $a\times c=2\times {{a}^{2}}\times {{b}^{2}}$
We have to look for the number whose product in $2{{a}^{2}}{{b}^{2}}$ but sum (ii) $-3ab$ if we look closely then we get $-2ab\,\,\,and\,\,-ab$ its product is $2{{a}^{2}}{{b}^{2}}$
And their sum is $-2ab-ab=-3ab$. So, we use these $2$ to split our middle term so,
${{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0$
${{a}^{2}}{{x}^{2}}+(-2ab-ab)x+2{{b}^{2}}=0$
Opening brackets, we get
${{a}^{2}}x-2abx-abx+2{{b}^{2}}=0$
Taking common in first two terms and last two terms we get,
\[ax(ax-2b)-b(ax-2b)=0\]
$(ax-b)(ax-2b)=0$
Now as product of two terms is zero, so it means either one of them is zero so,
$ax-b=0\,\,\,\,\,or\,\,\,\,\,\,ax-2b=0$
Simplifying we get,
\[x=\dfrac{b}{a}\,\,\,\,\,\,or\,\,\,\,\,\,x=\dfrac{2b}{a}\]
So, solution of ${{a}^{2}}{{x}^{2}}+2{{b}^{2}}=0$ and
$x=\dfrac{b}{a}\,\,\,\,\,and\,\,\,\,x=\dfrac{2b}{a}$

Note: We can check our solution, just to verify our answer, to do we will put $x=\dfrac{b}{a}\,\,\,\,\,and\,\,\,\,x=\dfrac{2b}{a}$ in ${{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0$ one by one and see if they satisfy equation or not putting $x=\dfrac{b}{a}$in ${{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0$ we get
\[{{a}^{2}}{{\left( \dfrac{b}{a} \right)}^{2}}-3ab\left( \dfrac{b}{a} \right)+2{{b}^{2}}=0\]
Simplifying we get
${{a}^{2}}\times \dfrac{{{b}^{2}}}{{{a}^{2}}}-3ab\times \dfrac{b}{a}+2{{b}^{2}}=0$
So, we get after cancelling like terms,
${{b}^{2}}-3{{b}^{2}}+2{{b}^{2}}=0$
So, $0=0$
Hence, $x=\dfrac{b}{a}$ is the correct solution.
Now putting $x=\dfrac{2b}{a}$ our equation we get
\[{{a}^{2}}{{\left( \dfrac{b}{a} \right)}^{2}}-3ab\left( \dfrac{b}{a} \right)+2{{b}^{2}}=0\]
Simplifying we get,
$4{{b}^{2}}-6{{b}^{2}}+2{{b}^{2}}=0$
Adding like terms, we get $0=0$
So, $x=\dfrac{2b}{a}$ satisfies our equation.
So, $x=\dfrac{2b}{a}$ is also the correct solution.