Question

How do you solve \[2{{x}^{2}}-5x+1=0\]?

Answer 1

Hint: The degree of the equation is the highest power to which the variable is raised. We can find whether the equation is linear, quadratic, cubic, etc. from the degree of the equation. For any quadratic equation \[a{{x}^{2}}+bx+c=0\], here \[a,b\]and \[c\in \]Real numbers. The roots of the equation can be found by using the formula method, which states that, the roots of the quadratic equation with real coefficients are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step answer:
We are asked to solve the given equation \[2{{x}^{2}}-5x+1=0\]. It means we have to find the roots of the given equation. The highest power to which x is raised is 2, so the degree of the equation is 2. It means that the equation is quadratic. We know that for a general quadratic equation \[a{{x}^{2}}+bx+c=0\], here \[a,b\]and \[c\in \]Real numbers. Using the formula method, the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], comparing the given equation with the general form. We get \[a=2,b=-5\And c=1\]. Substituting these values in the above formula we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 2 \right)\left( 1 \right)}}{2\times \left( 2 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-8}}{4} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{17}}{4} \\
\end{align}\]
Hence the roots of the given quadratic equation \[2{{x}^{2}}-5x+1=0\] are \[x=\dfrac{5+\sqrt{17}}{4}\]or \[x=\dfrac{5-\sqrt{17}}{4}\].

Note:
There are many ways to find the roots of a quadratic equation like the factorization method, completing the square method, formula method. We can find the roots of the quadratic equation by using any of these three methods. But, the factorization method should be preferred because it also tells us about whether roots are real or they are imaginary numbers.