Question

If the value of \[{{a}^{2}}+{{b}^{2}}=41\] and \[ab=4\], then the value of \[a+b\] is:
a)6
b)5
c)7
d)8

Answer 1

Hint: Represent \[{{a}^{2}}+{{b}^{2}}\] in the form of \[ab\] and \[a+b\]. Find the value of \[{{a}^{2}}+{{b}^{2}}\] from there.

Complete step-by-step answer:
It is given that \[{{a}^{2}}+{{b}^{2}}=41\] and \[ab=4\]. We have to find the value of \[a+b\] from there. Now, we will represent \[{{a}^{2}}+{{b}^{2}}\] in the form of \[ab\] and \[a+b\] in order to get the value of \[a+b\] from there.
We can write, \[{{a}^{2}}+{{b}^{2}}=({{a}^{2}}+{{b}^{2}}+2ab)-2ab\]
Now from basic algebra principles we know that \[{{a}^{2}}+{{b}^{2}}+2ab\] this quantity represents the value \[{{(a+b)}^{2}}\].
Therefore, we can write this as \[{{a}^{2}}+{{b}^{2}}=({{a}^{2}}+{{b}^{2}}+2ab)-2ab={{(a+b)}^{2}}-2ab\]
\[\Rightarrow {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
Now we know the values of \[{{a}^{2}}+{{b}^{2}}\] and \[ab\] both. Therefore, if we put these two values in the above equation we will get our desired answer.
Hence,\[\Rightarrow {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab=41+2\cdot 4\] = 41 + 8 = 49
Now we will have to take out the square root on both sides. It is clear that 49 can be written as \[{{7}^{2}}\].
Therefore,\[{{(a+b)}^{2}}={{7}^{2}}\]
Now, performing square root operation on both sides we get two distinct equations one with a positive root and another with a negative root. Because both a positive and negative value when squared generates a positive value.
So, we get \[(a+b)=7\] and \[(a+b)=-7\] respectively.
Hence, the correct answer to this question is option (c) 7.

Note: It is given that \[ab=4\]. Therefore, as the product is positive we can say that both a and b will have the same sign. Therefore, either both are positive or both are negative.