# Solve $x+\dfrac{1}{x} \,=\, 2.5$ by factorization method

$x+\dfrac{1}{x} \,=\, 2.5$ is an algebraic equation in terms of variable $x$. It is required to solve this algebraic equation by the factorization method.

### Simplifying the algebraic equation

The right hand side of equation is a decimal and write it in fraction form. It helps us to simplify the equation easily.

$\implies$ $x+\dfrac{1}{x} \,=\, \dfrac{5}{2}$

$\implies$ $\dfrac{x^2+1}{x} \,=\, \dfrac{5}{2}$

$\implies$ $2(x^2+1) \,=\, 5x$

$\implies$ $2x^2+2 \,=\, 5x$

$\implies$ $2x^2+2-5x \,=\, 0$

$\implies$ $2x^2-5x+2 \,=\, 0$

### Use factoring method to solve quadratic equation

It is a quadratic equation and it can be solved in four different ways but factorisation method is a simple way to find the solution for the quadratic equation $2x^2-5x+2 \,=\, 0$.

In this example, the literal coefficient of $x^2$ is $2$ and constant term is $2$. The product of them is equal to $4$ and it can be written as $4$ times $1$ and the sum of them is equal to $5$, which is literal coefficient of $x$. Hence, the quadratic equation can be solved by factoring method.

$\implies$ $2x^2-4x-x+2 \,=\, 0$

$\implies$ $2x \times x-2 \times 2x-x+2 \,=\, 0$

$2x$ is a common factor in the first two terms of the algebraic equation. So, take $2x$ common from them.

$\implies$ $2x(x-2)-x+2 \,=\, 0$

The third and fourth terms in the expression can be written in the following way to obtain $x-2$ common from them.

$\implies$ $2x(x-2)-1 \times x+ (-1) \times (-2) \,=\, 0$

$-1$ is a common factor in the third and fourth terms and take $-1$ common from them.

$\implies$ $2x(x-2)-1(x-2) \,=\, 0$

Now, $x-2$ is a common factor in the two terms. It can be taken common from them to get the left hand side expression in factoring form.

$\implies$ $(x-2)(2x-1) \,=\, 0$

### Get solutions of the quadratic equation

The product of two factors is equal to zero. According to zero product property, at least one of them is equal to zero.

$x-2 \,=\, 0 \,\,\,$ (or) $\,\,\, 2x-1 \,=\, 0$

$\implies$ $x \,=\, 2 \,\,\,$ (or) $\,\,\, 2x \,=\, 1$

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, 2 \,\,\,$ (or) $\,\,\, x \,=\, \dfrac{1}{2}$

Therefore $x = 2$ or $x = \dfrac{1}{2}$ is the solution for the quadratic equation by the factorization method.

### Verify the Roots of the Equation

Now, substitute both roots in the left hand side of the equation and check the value of the right hand side.

$(1) \,\,\,\,\,\,$ If $x \,=\, 2$

Substitute $x$ is equal to $2$ and evaluate the algebraic expression.

$x+\dfrac{1}{x} \,=\, 2+\dfrac{1}{2}$

$\implies$ $x+\dfrac{1}{x} \,=\, 2+0.5$

$\,\,\, \therefore \,\,\,\,\,\,$ $x+\dfrac{1}{x} \,=\, 2.5$

$(2) \,\,\,\,\,\,$ If $x \,=\, \dfrac{1}{2}$

Similarly, substitute $x$ is equal to $\dfrac{1}{2}$ and find the value of algebraic expression.

$x+\dfrac{1}{x} \,=\, \dfrac{1}{2}+\dfrac{1}{\dfrac{1}{2}}$

$\implies$ $x+\dfrac{1}{x} \,=\, 0.5+2$

$\,\,\, \therefore \,\,\,\,\,\,$ $x+\dfrac{1}{x} \,=\, 2.5$

Therefore, it is verified that $x = 2$ is true and $x = \dfrac{1}{2}$ is also true.

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