Question 12 Marks
Write the solution of inequality $x^2-4 x+3 \geq 0$.
Answer
$\begin{array}{l}x^2-4 x+3 \geq 0 \\\Rightarrow \quad(x-3)(x-1) \geq 0\end{array}$
Hence, two following cases are possible :
(i) $x-3 \geq 0$ and $x-1 \geq 0$
(ii) $x-3 \leq 0$ and $x-1 \leq 0$
$\text {for case (i)}\quad x-3 \geq 0 \Rightarrow x \geq 3$
$\text {and}\quad x-1 \geq 1 \Rightarrow x \geq 1$
So the possible solution of above case is $x \geq 3$
Hence, the required solution of given inequality.
$\begin{array}{l}x \geq 3 \text { and } x \leq 1 \\ (-\infty, 1] \cup[3, \infty)\end{array}$
View full question & answer→$\begin{array}{l}x^2-4 x+3 \geq 0 \\\Rightarrow \quad(x-3)(x-1) \geq 0\end{array}$
Hence, two following cases are possible :
(i) $x-3 \geq 0$ and $x-1 \geq 0$
(ii) $x-3 \leq 0$ and $x-1 \leq 0$
$\text {for case (i)}\quad x-3 \geq 0 \Rightarrow x \geq 3$
$\text {and}\quad x-1 \geq 1 \Rightarrow x \geq 1$
So the possible solution of above case is $x \geq 3$
Hence, the required solution of given inequality.
$\begin{array}{l}x \geq 3 \text { and } x \leq 1 \\ (-\infty, 1] \cup[3, \infty)\end{array}$