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Chapter 2: Problem 80

Explain why $$ x^{2}+4 x+y^{2}-10 y \geq-29 $$ for all real numbers \(x\) and \(y\).

Short Answer

Expert verified
The inequality \(x^2+4x+y^2-10y \geq -29\) holds for all real numbers x and y because when we complete the square for both x and y terms, we get \((x + 2)^2 + (y - 5)^2 \geq 0\). Since the sum of two square terms is always non-negative, the left side of the inequality is always greater than or equal to -29.

Step by step solution

01

Complete the square for both x and y

We rewrite the inequality as a sum of squares for both x and y terms. $$ x^2 + 4x + y^2 - 10y \geq -29 $$ To complete the square, we add and subtract the square of half of the coefficients of the linear terms (4 and -10). $$ (x^2 + 4x + 4) + (y^2 - 10y + 25) \geq -29 + 4 + 25 $$ Now, we can write the left side of the inequality as the sum of two perfect squares: $$ (x + 2)^2 + (y - 5)^2 \geq 0 $$
02

Simplify the inequality

Since \( (x+2)^2 \) and \( (y-5)^2 \) are both square terms, they are always non-negative (greater than or equal to 0). The sum of two non-negative numbers is always non-negative: $$ (x + 2)^2 + (y - 5)^2 \geq 0 $$
03

Compare with -29

Since we already have the inequality: $$ (x + 2)^2 + (y - 5)^2 \geq 0 $$ It is clear that the left side is always greater than or equal to 0. Therefore: $$ (x + 2)^2 + (y - 5)^2 \geq -29 $$ The inequality holds for all real numbers x and y, showing that the original inequality is true.

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