Question

Solve the following by factorization method:

$\frac{2}{x^2}-\frac{5}{x}+2=0;x\ne 0$

Answer 1

Step 1:

Express the equation in the standard quadratic form:

Simplify the given equation.

$\begin{array}{rcl}\frac{2}{x^2}-\frac{5}{x}+2& =& 0\\ & \Rightarrow & \frac{2-5x+2x^2}{x^2}=0\\ & \Rightarrow & 2-5x+2x^2=0\\ & \Rightarrow & 2x^2-5x+2=0\end{array}$ as x is not equal to zero.

Step 2:

Split the middle term and factorize the equation :

Find two numbers such that their sum is equal to the coefficient of $x$, that is $-5$.

And their product is equal to the product of the coefficient of $x^2$ and the constant term, that is $2\times 2=4$.

Notice that,

$\begin{array}{rcl}\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{numbers}& =& \left(\left(-4\right)+\left(-1\right)\right)\\ & =& -5\end{array}$

$\begin{array}{rcl}\mathrm{Product}\mathrm{of}\mathrm{two}\mathrm{numbers}& =& \left(-4\right)\times \left(-1\right)\\ & =& 4\end{array}$

Therefore, the two numbers are $-4$ and $-1$.

Express the quadratic equation as a product of its factors.

$\begin{array}{rcl}2x^2-5x+2& =& 0\\ & \Rightarrow & 2x^2-4x-1x+2=0\\ & \Rightarrow & 2x\left(x-2\right)-1\left(x-2\right)=0\\ & \Rightarrow & \left(2x-1\right)\left(x-2\right)=0\\ & \Rightarrow & 2x-1=0\mathrm{or}x-2=0\\ & \Rightarrow & x=\frac{1}{2}\mathrm{or}x=2\end{array}$

So, the roots of the given quadratic equation are $x=\frac{1}{2}$ and $x=2$.