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Solving inequalities
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Solving inequalities

Problem 1.

Prove that a2 - 6a + 10  > 0 for any value of a

Solution

Transform the original inequality as
a2 - 6a + 9 + 1 > 0
then a2 - 6a + 9  can be represented as (a-3)2 that is
( a - 3 )2 + 1 > 0
( a - 3 )2 > -1
since the expression in brackets will be positive or equal to zero when squared,
then the inequality will be satisfied for any value of a.

Problem 2.

Prove that for any value of a the inequality holds
a(a-8) > 2(a-13)

Solution

  1. Let's transform the original inequality
    a(a-8) - 2(a-13) > 0
    a2- 8a - 2a + 26 > 0
    a2 - 10a + 26 > 0
  2. represent the left-hand side of the inequality as
    a2-10a+25+1>0
    (a-5)2+1>0
  3. (a-5)2>-1
since the expression in brackets will be positive or equal to zero when squared,
then the inequality will hold for any value of a.

Problem 3. Fractional rational inequality

Solve the inequality

( x−2 ) / ( x+3 ) >0
Solution:

  1. Find the zeros of the numerator and denominator:
    x−2 = 0 ⇒ x = 2
    x+3 = 0 ⇒ x = −3
  2. Divide the number line into intervals:
    ( −∞, −3 )
    ( −3, 2 )
    ( 2, ∞ )
  3. Determine the signs on each interval by substituting trial values:
    For x = −4
    in the expression ( x−2 ) / ( x+3 ) → (−4−2 ) / (−4+3 ) = − 6 / −1 = 6 (positive)
    For x=0
    ( 0 − 2 ) / ( 0 + 3 ) = −2 / 3 ​ (negative)
    For x=3
    ( 3 − 2 ) / ( 3 + 3 ) = 1 / 6​ (positive)

The inequality ( x − 2 ) / ( x + 3 ) > 0 is satisfied for x∈(−∞,−3)∪(2,∞)

Answer: (−∞,−3)∪(2,∞)

Problem 4. Modulus Inequality

Solve the inequality

| x - 4 | < 5
Solution:

By definition of modulus: −5 < x−4 < 5
Add 4 to all parts: −1 < x < 9

 Inequalities | Описание курса | Differential calculus