Higher Engineering Mathematics

16 NUMBER AND ALGEBRA

From the general rule stated above in equation (2):

ifx^2 <4 then−

√

4 <x<

√

4

i.e. the inequality:x^2 <4 is satisfied when:

− 2 <x< 2

Problem 14. Solve the inequality:

(y−3)^2 ≤ 16

From equation (2), −

√

16 ≤(y−3)≤

√

16

i.e. − 4 ≤(y−3)≤ 4

from which, 3 − 4 ≤y≤ 4 +3,

i.e. − 1 ≤y≤ 7

Now try the following exercise.

Exercise 11 Further problems on inequali-

ties involving square functions

Solve the following inequalities:

1. z^2 > 16 [z>4orz<−4]


2. z^2 < 16 [− 4 <z<4]


3. 2x^2 ≥6[x≥

√

3orx≤−

√

3]

4. 3k^2 − 2 ≤ 10 [− 2 ≤k≤2]


5. (t−1)^2 ≤ 36 [− 5 ≤t≤7]


6. (t−1)^2 ≥ 36 [t≥7ort≤−5]


7. 7− 3 y^2 ≤−5[y≥2ory≤−2]


8. (4k+5)^2 > 9

[

k>−

1

2

or k<− 2

]

2.6 Quadratic inequalities

Inequalities involving quadratic expressions are

solved using eitherfactorizationor ‘completing the

square’. For example,

x^2 − 2 x−3 is factorized as (x+1)(x−3)

and 6x^2 + 7 x−5 is factorized as (2x−1)(3x+5)

If a quadratic expression does not factorize, then

the technique of ‘completing the square’ is used. In

general, the procedure forx^2 +bx+cis:

x^2 +bx+c≡

(

x+

b

2

) 2

+c−

(

b

2

) 2

For example,x^2 + 4 x−7 does not factorize; com-

pleting the square gives:

x^2 + 4 x− 7 ≡(x+2)^2 − 7 − 22 ≡(x+2)^2 − 11

Similarly,

x^2 − 6 x− 5 ≡(x−3)^2 − 5 − 32 ≡(x−3)^2 − 14

Solving quadratic inequalities is demonstrated in the

following worked problems.

Problem 15. Solve the inequality:

x^2 + 2 x− 3 > 0

Since x^2 + 2 x− 3 >0 then (x−1)(x+3)>0by

factorizing. For the product (x−1)(x+3) to be

positive,

either (i) (x−1)> 0 and (x+3)> 0

or (ii) (x−1)<0 and (x+3)< 0

(i) Since (x−1)>0 thenx>1 and since (x+3)> 0

thenx>− 3

Both of these inequalities are satisfied only when

x> 1

(ii) Since (x−1)<0 thenx<1 and since (x+3)< 0

thenx<− 3

Both of these inequalities are satisfied only when

x<− 3

Summarizing, x^2 + 2 x− 3 >0 is satisfied when

eitherx>1orx<− 3

Problem 16. Solve the inequality:

t^2 − 2 t− 8 < 0

Since t^2 − 2 t− 8 <0 then (t−4)(t+2)<0by

factorizing.

For the product (t−4)(t+2) to be negative,

either (i) (t−4)> 0 and (t+2)< 0

or (ii) (t−4)< 0 and (t+2)> 0

(i) Since (t−4)>0 thent>4 and since (t+2)< 0

thent<− 2

It is not possible to satisfy botht>4 andt<−2,

thus no values oftsatisfies the inequality (i)

(ii) Since (t−4)<0 thent<4 and since (t+2)> 0

thent>− 2

Hence, (ii) is satisfied when− 2 <t< 4