If \(a^2+b^2+c^2=4.84\) and \(ab+bc+ca=2.7\), find \(a+b+c\), given that \(a+b+c>0\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3.2\)
The square of a sum of three terms is \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\). The square root gives two signs, and the condition \(a+b+c>0\) keeps only the positive one.
\((a+b+c)^2=4.84+2\times2.7=4.84+5.4=10.24\)
\(\sqrt{10.24}=3.2\), so \(a+b+c=3.2\) or \(-3.2\)
\(a+b+c>0\), so \(a+b+c=3.2\)
Why the other options are incorrect
Option (1) \(-3.2\) — Incorrect. The value \(-3.2\) is ruled out by the condition \(a+b+c>0\).
Option (2) \(\pm3.2\) — Incorrect. The value \(-3.2\) is ruled out by the condition \(a+b+c>0\), so only one value remains.
Option (4) \(2.2\) — Incorrect. \(2.2^2=4.84\): this leaves out the term \(2(ab+bc+ca)\); \((a+b+c)^2\) is 10.24, not 4.84.