If for real values of x, (x^{2} - 3x + 2 > 0) and (x^{2} - 3... | MATH - CONDITION FOR COMMON ROOTS, QUADRATIC EXPRESSIONS AND POSITION OF ROOTS | SolveeIt

If for real values of x, $x^{2} - 3x + 2 > 0$ and $x^{2} - 3x - 4 \leq 0$ ,

Then

$- 1 \leq x < 1$

$- 1 \leq x < 4$

$- 1 \leq x < 1$ or $2 < x \leq 4$

$2 < x \leq 4$

Answer

$- 1 \leq x < 1$ or $2 < x \leq 4$

Explanation

$x^{2} - 3x + 2 > 0$ or $(x - 1)(x - 2) > 0$

∴ $\mathbf{x}\mathbf{\in}\mathbf{(}\mathbf{- \infty}\mathbf{,1)}\mathbf{\cup}\mathbf{(2,}\mathbf{\infty}\mathbf{)}$ …..(i)

Again $x^{2} - 3x - 4 \leq 0$ or $(x - 4)(x + 1) \leq 0$

∴ $x \in \lbrack - 1,4\rbrack$ …..(ii)

From eq. (i) and (ii), $x \in \lbrack - 1,1) \cup (24\rbrack$ ⇒ $- 1 \leq x < 1$ or

$2 < x \leq 4$

Question: If for real values of x, \(x^{2} - 3x + 2 > 0\) and \(x^{2} - 3x - 4 \leq 0\), Then...