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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