Question

# Solve the quadratic equation $2{{x}^{2}}-x+\dfrac{1}{8}=0$ to find the value of x.

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Hint: For the quadratic equations containing rational terms, it is difficult to solve them using the method of splitting the middle term. For these type of quadratic equations, to find the value of x, we use quadratic formula from which, the roots for the quadratic equation $a{{x}^{2}}+bx+c=0$ is given by the formula x = $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Using this quadratic formula, we can solve this question.

In a quadratic equation, if we have a quadratic equation having rational terms, then these quadratic equations are solved by using the quadratic formula. For a quadratic equation $a{{x}^{2}}+bx+c=0$, the roots of the quadratic equation can be found by the formula,
x = $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . . . . . . . . . . . . . (1)
In this question, we are given a quadratic equation $2{{x}^{2}}-x+\dfrac{1}{8}=0$ and we have to solve this quadratic equation to find the value of x. Substituting a = 2, b = -1 and c = $\dfrac{1}{8}$ in formula (1), we get,
x = $\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 2 \right)\left( \dfrac{1}{8} \right)}}{2\left( 2 \right)}$
\begin{align} & \Rightarrow x=\dfrac{1\pm \sqrt{1-1}}{4} \\ & \Rightarrow x=\dfrac{1\pm \sqrt{0}}{4} \\ & \Rightarrow x=\dfrac{1}{4} \\ \end{align}
Hence, the value of x which satisfy the quadratic equation $2{{x}^{2}}-x+\dfrac{1}{8}=0$ is $\dfrac{1}{4}$.