# Prove that a^2-4a+b^2+10b+29>=0 if a,b are real numbers .

Asked on by maisaphie

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to prove that a^2 - 4a + b^2 + 10b + 29>=0, for real values of a and b.

a^2 - 4a + b^2 + 10b + 29

=> a^2 - 4a + 4 + b^2 + 10b + 25

=> (a - 2)^2 + (b + 5)^2

The sum of squares of real numbers is always positive.

This proves that a^2 - 4a + b^2 + 10b + 29 >= 0

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Alll we need to do is to express the left sides as a sum of postivie numbers.

We'll create perfect squares to the left side:

(a^2 - 4a + 4) + (b^2 + 10b + 25) - 4 - 25 + 29 >=0

We notice that we've added the numbers 4 and 25 to complete the squares. For the inequality to hold, we'll have to subtract these added values.

(a - 2)^2 + (b + 5)^2 - 29 + 29 >= 0

We'll eliminate like terms:

(a - 2)^2 + (b + 5)^2 > =0

Since the squares are always positive, the inequality is verified, for any real values of a and b.

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