When can \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\) generally not be considered true?
Answer and explanation
Correct answer: It is not a generally true rule
For non-negative \(a\) and \(b\), squaring the two sides gives \(a+b+2\sqrt{ab}\) on the left and \(a+b\) on the right. They are equal only when \(2\sqrt{ab}=0\), that is, when \(ab=0\). Hence, this is not a general rule; it works only in special cases. Option D describes one such special case, so it cannot represent the general conclusion. For example, \(\sqrt{4}+\sqrt{9}=5\), whereas \(\sqrt{4+9}=\sqrt{13}\). Exam tip: do not directly combine a sum of square roots into one square root.
Frequently asked questions
What is the correct answer to this question?
It is not a generally true rule
Why is this the correct answer?
For non-negative \(a\) and \(b\), squaring the two sides gives \(a+b+2\sqrt{ab}\) on the left and \(a+b\) on the right. They are equal only when \(2\sqrt{ab}=0\), that is, when \(ab=0\). Hence, this is not a general rule; it works only in special cases. Option D describes one such special case, so it cannot represent the general conclusion. For example, \(\sqrt{4}+\sqrt{9}=5\), whereas \(\sqrt{4+9}=\sqrt{13}\). Exam tip: do not directly combine a sum of square roots into one square root.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Irrational numbers.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
