Question

solve show your solution​

Answer 1

$\large\underline{\sf{Solution-1}}$

$\rm \: {x}^{2} + 9x + 14 > 0$

$\rm \: {x}^{2} + 7x + 2x + 14 > 0$

$\rm \: x(x + 7) + 2(x + 7) > 0$

$\rm \: (x + 7)(x + 2) > 0$

$\bf\implies \:x < - 7 \: \: or \: \: x > - 2$

$\large\underline{\sf{Solution-2}}$

$\rm \: {x}^{2} - 10x + 16 < 0$

$\rm \: {x}^{2} - 8x - 2x + 16 < 0$

$\rm \: x(x - 8) - 2(x - 8) < 0$

$\rm \: (x - 8)(x - 2) < 0$

$\bf\implies \:2 < x < 8$

$\large\underline{\sf{Solution-3}}$

$\rm \: {x}^{2} + 6x \geqslant - 5$

$\rm \: {x}^{2} + 6x + 5 \geqslant 0$

$\rm \: {x}^{2} + x + 5x + 5\geqslant0$

$\rm \: x(x + 1) + 5(x + 1) \geqslant 0$

$\rm \: (x + 1)(x + 5) \geqslant 0$

$\bf\implies \:x \leqslant - 5 \: \: or \: \: x \geqslant - 1$

$\large\underline{\sf{Solution-4}}$

$\rm \: {x}^{2} - 5x - 14 > 0$

$\rm \: {x}^{2} - 7x + 2x - 14 > 0$

$\rm \: x(x - 7) + 2(x - 7) > 0$

$\rm \: (x - 7)(x + 2) > 0$

$\bf\implies \:x < - 2 \: \: or \: \: x > 7$

$\large\underline{\sf{Solution-5}}$

$\rm \: {2x}^{2} + 11x + 12 < 0$

$\rm \: {2x}^{2} + 8x + 3x + 12 < 0$

$\rm \: 2x(x + 4) + 3(x + 4) < 0$

$\rm \: (x + 4)(2x + 3) < 0$

$\bf\implies \: - 4 < x < - \dfrac{3}{2}$

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Basic Concept Used

If a and b are two positive real numbers such that a < b,

$\boxed{\tt{ (x - a)(x - b) < 0 \: \rm\implies \: \: a < x < b \: }}$

$\boxed{\tt{ (x - a)(x - b) \leqslant 0 \: \rm\implies \: \: a \leqslant x \leqslant b \: }}$

$\boxed{\tt{ (x - a)(x - b) > 0 \: \: \rm\implies \:x < a \: \: or \: \: x > b \: }}$

$\boxed{\tt{ (x - a)(x - b) \geqslant 0 \: \: \rm\implies \:x \leqslant a \: \: or \: \: x \geqslant b \: }}$

Answer 2

Answer:

1. x > -2 or x < -7

2. 2 < x < 8

3. x ≥ -1 or x ≤ -5

4. x > 7 or x < -2  

5. -4 < x < -1,5

Step-by-step explanation:

1.

x² + 9x + 14 = (x + 2)(x + 7) > 0

⇒ x + 2 > 0, x + 7 > 0 ↔ x > -2, x > -7 ⇒ x > -2

or x + 2 < 0, x + 7 < 0 ↔ x < -2, x < -7 ⇒ x < -7

2.

x² - 10x + 16 = (x - 2)(x - 8) < 0

x - 2 > x - 8 ⇒ x - 2 > 0, x - 8 < 0

⇔ 2 < x < 8

3.

x² + 6x ≥ -5

x² + 6x + 5 = (x + 1)(x + 5) ≥ 0

⇒ x + 1 ≥ 0, x + 5 ≥ 0 ↔ x ≥ -1, x ≥ -5 ⇒ x ≥ -1

or x + 1 ≤ 0, x + 5 ≤ 0 ↔ x ≤ -1, x ≤ -5 ⇒ x ≤ -5

4.

x² - 5x - 14 = (x + 2)(x - 7) > 0

⇒ x + 2 > 0, x - 7 > 0 ↔ x > -2, x > 7 ⇒ x > 7

or x + 2 < 0, x - 7 < 0 ↔ x < -2, x < 7 ⇒ x < -2

5.

2x² + 11x + 12 = (x + 4)(2x + 3) < 0

⇒ x + 4 < 0, 2x + 3 > 0 ↔ x < -4, x > -1.5 (false)

or x + 4 > 0, 2x + 3 < 0 ↔ x > -4, x < -1.5 ⇒ -4 < x < -1.5