Question

Solve – x² + 3x – 2 ≥ 0

Answer 1

The given inequality is

– x² + 3x – 2 ≥ 0

x² – 3x + 2 < 0   ......(1)

x² – 3x + 2 = x² – 2x – x + 2

= x(x – 2) – 1(x – 2)

x² – 3x + 2 = (x – 1) (x – 2)  ......(2)

The critical numbers are

x – 1 = 0 or x – 2 = 0

The critical numbers are

x = 1 or x = 2

Divide the number line into three intervals

`(– oo, 1)`, (1, 2) and `(2, oo)`.

Interval	Sign of x – 1	Sign of x – 2	Sign of x2 – 3 + 2
`(– oo, 1)`	–	–	+
(1, 2)	+	–	–
`(2, oo)`	+	+	+

(i) `(– oo, 1)`

When x < 1 say x = 0

The factor x – 1 = 0 – 1 = – 1 < 0 and

x – 2 = 0 – 2 = – 2 < 0

x – 1 < 0 and x – 2 < 0

⇒ (x – 1)(x – 2) > 0

Using equation (2) x² – 3x + 2 > 0

∴ The inequality x² – 3x + 2 ≤ 0 is not true in the interval (– ∞, 1)

(ii) (1, 2)

When x lies between 1 and 2 say x = `3/2`

The factor x – 1 = `3/2 - 1 = 1/2 > 0` and

x – 2 = `3/2 - 2 = - 1/2 - < 0`

x – 1 > 0 and x – 2 < 0

⇒ (x – 1)(x – 2) < 0

Using equation (2) x² – 3x + 2 < 0

∴ The inequality x² – 3x + 2 ≤ 0 is true in the interval (1, 2)

(iii) `(2, oo)`

When x > 2 say x = 3

The factor x – 1 = 3 – 1 = 2 > 0 and

x – 2 = 3 – 2 = 1 > 0

x – 1 > 0 and x – 2 > 0

= (x – 1)(x – 2) > 0

Using equation (2) x² – 3x + 2 > 0

∴ The inequality x² – 3x + 2 ≤ 0 is not true in the interval (2, ∞)

We have proved the inequality x² – 3x + 2 ≤ 0 is true in the interval [1, 2].

But it is not true in the interval

`(– oo, 1)` and `(2, oo)`

∴ The solution set is [1, 2]