⇒ (x + 3) (2x – 1) (x – 1) (x – 2) ≤ 0

⇒ The critical points are –3, 1/2,  1 and 2

When x = 0, the inequality is true

The solution set is [–3, 1/2] ∪ (1, 2)

Equal roots, if D = 0 ⇒ B² – 4AC = 0

(2mc)² – 4(1 + m²)(c² – a²) = 0

            4m²c² – 4[c² + m²c² – m²a² – a²] = 0

As                    4 ≠ 0

            c² – m²a² – a² = 0 ⇒ c² = a²(1 + m²)

When x < - 2, |x – 1| > 3

So, |x + 2| + |x – 1| > 3

When x > 1, |x + 2| > 3 and so |x + 2| + |x – 1| > 3.

When -2 < × < 1, |x + 2| + |x – 1| = 3

Hence there is no real number x such that |x + 2| + |x – 1| < 2.

We know that

∴ |2x – 3| > 0 for every real value of x.

∴ |2x – 3| + 17 > 17

∴ the minimum value of |2x + 3| + 17 is 17.

Given |3x + 7| > 12

3x + 7 < – 12 or 3x + 7 > 12

3x < – 19 or 3x > 5

x < – 19 / 3 or x > 5 / 3

∴ the solution set is 

A         2x² – 13x + 21 = 0 ⇒ 2x² – 7x – 6x + 21 = 0

⇒        x(2x – 7) – 3(2x – 7) = 0 ⇒ (2x – 7)(x – 3) = 0

⇒        x = 7/2 or x = 3

B         2x² – 9x + 7 = 0 ⇒ (2x – 7)(x – 1) = 0

⇒        x = 7 / 2 or x = 1

C         x² – 6x + 9 = 0

⇒        (x – 3)(x – 3) = 0

⇒        x = 3 or x = 3

D         2x² – 21x + 49 = 0

⇒        (2x – 7)(x – 7) = 0

⇒        x = 7 / 2 or x = 7

Given |x + 3| < 5

⇒ – 5 < x + 3 < 5

⇒ – 5  – 3 < x < 5 – 3

⇒ – 8 < x < 2

∴ The integers which satisfy the given in equation are –8, –7, –6, –5, –4, –3, –2, –1, 0, 1 and 2.

Hence the required number of integers is 11.

Given x = |a| b and |b| > 2

a + xb = a + |a| b²

and a – xb = a - |a|b²

As|–6| > 1, the magnitude of |a| b² is greater than that of a.

The sign of either expression is determined by the sign of this term.

i.e. a + |a| b² > 0 and a – |a| b² < 0

The equality holds when a = 0.

Given, |9 – x| < 2 – 3x

Let x < 9

Then, 9 – x < 2 – 3x ⇒ 2x < – 7

⇒ x < – 7 / 2 which agrees with x < 9

Let x > 9

Then –(9 – x) < 2 – 3x

– 9 + x < 2 – 3x

4x < 11

 which does not agree with x > 9.

∴ the solution set is .

Since, – 1.5 is a root of ax² + x – 3 = 0

∴         a(–1.5)² + (–1.5) – 3 = 0

⇒        2.25 a – 4.5 = 0 ⇒ a =