Higher Engineering Mathematics

INEQUALITIES 17

A

Summarizing, t^2 − 2 t− 8 <0 is satisfied when
− 2 <t< 4

Problem 17. Solve the inequality:

x^2 + 6 x+ 3 < 0

x^2 + 6 x+3 does not factorize; completing the
square gives:

x^2 + 6 x+ 3 ≡(x+ 3 )^2 + 3 − 32

≡(x+ 3 )^2 − 6

The inequality thus becomes:(x+ 3 )^2 − 6 <0or

(x+ 3 )^2 < 6

From equation (2),−

√

6 <(x+3)<

√

6

from which, (−

√

6 −3)<x<(

√

6 −3)

Hence,x^2 + 6 x+ 3 <0 is satisfied when
−5.45<x<−0.55correct to 2 decimal places.

Problem 18. Solve the inequality:

y^2 − 8 y− 10 ≥ 0

y^2 − 8 y−10 does not factorize; completing the
square gives:

y^2 − 8 y− 10 ≡(y−4)^2 − 10 − 42

≡(y−4)^2 − 26

The inequality thus becomes:(y− 4 )^2 − 26 ≥0or
(y− 4 )^2 ≥ 26

From equation (1), (y−4)≥

√

26 or (y−4)≤−

√

26

from which, y≥ 4 +

√

26 ory≤ 4 −

√

26

Hence,y^2 − 8 y− 10 ≥0 is satisfied wheny≥9.10

ory≤−1.10correct to 2 decimal places.

Now try the following exercise.

Exercise 12 Further problems on quadratic

inequalities

Solve the following inequalities:

1.x^2 −x− 6 >0[x>3orx<−2]

2.t^2 + 2 t− 8 ≤0[− 4 ≤t≤2]

3. 2x^2 + 3 x− 2 < 0

[

− 2 <x<

1

2

]

4.y^2 −y− 20 ≥0[y≥5ory≤−4]

5.z^2 + 4 z+ 4 ≤4[− 4 ≤z≤0]

6.x^2 + 6 x+ 6 ≤ 0

[(−

√

3 −3)≤x≤(

√

3 −3)]

7.t^2 − 4 t− 7 ≥ 0

[t≥(

√

11 +2) ort≤(2−

√

11)]

8.k^2 +k− 3 ≥ 0

[

k≥

(√

13

4

−

1

2

)

ork≤

(

−

√

13

4

−

1

2

)]