Question

prove that a^2 - 6a + 9 + b^2 ≥ 0 for all real values of a and b (using algebra)

Answer 1

Answer: Proof given below...

Step-by-step explanation:

a^2 - 6a + 9 + b^2

= [a^2 - 2×3×a + 3^2] + b^2

= (a- 3)^2+ b^2

As, square of any number is always positive

So, (a- 3)^2 ≥0, and b^2≥ 0

Hance, a^2 - 6a + 9 + b^2 ≥ 0

Answer 2

$\Large\frak{\underline{\underline{Question:}}}$

prove that a^2 - 6a + 9 + b^2 ≥ 0 for all real values of a and b (using algebra)

$\Large\frak{\underline{\underline{Answer:}}}$

a^2 - 6a + 9 + b^2

= [a^2 - 2×3×a + 3^2] + b^2

= (a- 3)^2+ b^2

As, square of any number is always positive

So, (a- 3)^2 ≥0, and b^2≥ 0

Hence, a^2 - 6a + 9 + b^2 ≥ 0