Question

find the value of a so that expression x^2-(a+2)x+4 is always positive​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that,

$\rm :\longmapsto\: {ax}^{2} + bx + c > 0 \: iff \: a > 0 \: and \: {b}^{2} - 4ac < 0$

Now,

Given that,

The quadratic expression is

$\rm :\longmapsto\:f(x) = {x}^{2} - (a + 2)x + 4$

On comparing with

$\rm :\longmapsto\: {ax}^{2} + bx + c, \: we \: get$

$\red{\rm :\longmapsto\:a = 1}$

$\red{\rm :\longmapsto\:b = - (a + 2)}$

$\red{\rm :\longmapsto\:c = 4}$

Since,

$\rm :\longmapsto\:a = 1 > 0$

and

To the expression,

$\rm :\longmapsto\:f(x) = {x}^{2} - (a + 2)x + 4 > 0$

we must have

$\rm :\longmapsto\: {\bigg( - (a + 2)\bigg) }^{2} - 4 \times 4 \times 1 < 0$

$\rm :\longmapsto\: {a}^{2} + 4 + 4a - 16 < 0$

$\rm :\longmapsto\: {a}^{2} + 4a - 12 < 0$

$\rm :\longmapsto\: {a}^{2} + 6a - 2a - 12 < 0$

$\rm :\longmapsto\:a(a + 6) - 2(a + 6) < 0$

$\rm :\longmapsto\:(a + 6)(a - 2) < 0$

$\bf\implies \:a \: \in \: (-6, 2)$

Additional Information :-

Let a and b are real numbers such that a > b, then

$\rm :\longmapsto\:(x - a)(x - b) < 0 \implies \: b < x < a$

$\rm :\longmapsto\:(x - a)(x - b) \leqslant 0 \implies \: b \leqslant x \leqslant a$

$\rm :\longmapsto\:(x - a)(x - b) > 0 \implies \: \in \: ( \infty ,b) \: \cup \: (a, \infty )$

$\rm :\longmapsto\:(x - a)(x - b) \geqslant 0 \implies \: \in \: ( \infty ,b] \: \cup \: [a, \infty )$

$\rm :\longmapsto\:(x + a)(x + b) < 0 \implies \: - a < x < - b$

$\rm :\longmapsto\:(x + a)(x + b) \leqslant 0 \implies \: - a \leqslant x \leqslant - b$

Answer 2

Answer:

is it previous class questions?