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Find the solution set of $x^{2}-3 x+2<0$.
Given :
The given equation is $x^{2}-3 x+2<0$.
To do :
We have to find the solution set of the given equation.
Solution :
Roots of a quadratic equation $ax^2 + bx + c$ is given by,
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
For, $x^{2}-3 x+2<0$
$x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(2)}}{2(1)}$
$x = \frac{3 \pm \sqrt{9-8}}{2}$
$x = \frac{3 \pm 1}{2}$
$x = \frac{3+1}{2}$ or $x = \frac{3-1}{2}$
$x = 2$ or $x = 1$
Therefore, $(x-2)(x-1)=0$
$(x-2)(x-1)<0$
Either $(x-2)$ or $(x-1)$ should be negative to satisfy the above statement.
Therefore,
$x-2>0$ and $x-1<0$ or $x-2<0$ and $x-1>0$
$x>2$ and $x<1$ or $x<2$ and $x>1$
$x>2$ and $x<1$ is not possible.
So, $x<2$ and $x>1$.
Therefore, the solution set is $x \in (1, 2)$, Here 1 and 2 are not included.
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