Question

Solve the given quadratic equation: ${{x}^{2}}-3x+2=0$

Answer 1

Hint: Consider the given quadratic equation ${{x}^{2}}-3x+2=0$ and write it as, ${{x}^{2}}-x-2x+2=0$ then factorize it in terms of $x$ and finally get values for $x$.
We are given a quadratic equation ${{x}^{2}}-3x+2=0$ and we have to solve to find values of $x$.

Complete step-by-step answer:
${{x}^{2}}-3x+2=0$is considered as a quadratic equation. The general form of quadratic equation is $a{{x}^{2}}+bx+c=0$
Here $x$represents unknown value and a, b, c are known numbers where $a\ne 0$ otherwise it becomes linear due to absence of ${{x}^{2}}$ term. The numbers a, b, c are coefficient, the linear coefficient and the constant or free term.
The values of $x$ that satisfy the equations are called solutions of the equation or roots of the quadratic equation. A quadratic equation has at most two solutions. If there is no real solution then there are two complex solutions/roots. If there is only one solution, one says that it is a double root.
 A quadratic equation of form $a{{x}^{2}}+bx+c=0$ can be factored as $\left( x-r \right)\left( x-s \right)=0$where r and s are solutions of $x$.
The quadratic equation only contains power of $x$ that are non negative integers and therefore it is a polynomial equation. In particular it is a second degree polynomial equation.
As the equation given is ${{x}^{2}}-3x+2=0$, by splitting the middle term, we get
$\begin{align}
  & {{x}^{2}}-x-2x+2=0 \\
 & \Rightarrow \left( x-1 \right)\left( x-2 \right)=0 \\
\end{align}$
So for the given equation 1 and 2 satisfies as the solution or roots.
Therefore the solution for the given quadratic equation is $x=1,2$ .

Note: We can also solve this method by a formula or by Sridhar acharya formula which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for quadratic equation $a{{x}^{2}}+bx+c=0$.