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# The product of two numbers is 120 and the sum of their squares is 289. The sum of the numbers is:a) 20b) 23c) 16d) None of these

Last updated date: 02nd Aug 2024
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Hint: Let two numbers be ‘a’ and ‘b’. As it is given, that product of two numbers is 120, therefore, we have $a\times b=120$. Also, it is said that the sum of squares of those two numbers is 289. So, we can write: ${{a}^{2}}+{{b}^{2}}=289$.
To find the sum of ‘a’ and ‘b’ use the identity: ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Get the values of $\left( a\times b \right)$and ${{a}^{2}}+{{b}^{2}}$ , then substitute in the identity to get the value of the sum of two numbers.

Let us consider two numbers ‘a’ and ‘b’.
We have been given that the product of two numbers is 120;
i.e. $a\times b=120......(1)$
Also, we have the sum of their squares given as 289,
Therefore, ${{a}^{2}}+{{b}^{2}}=289......(2)$

We need to find the sum of ‘a’ and ‘b’ i.e. $a+b$
By using the identity: ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
We get: $\left( a+b \right)=\sqrt{{{a}^{2}}+2ab+{{b}^{2}}}......(3)$
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Substitute the value of $\left( a\times b \right)$ and $\left( {{a}^{2}}+{{b}^{2}} \right)$ from equation (1) and (2) in equation (3);
We get:
\begin{align} & \left( a+b \right)=\sqrt{289+\left( 2\times 120 \right)} \\ & =\sqrt{289+240} \\ & =\sqrt{529} \\ & =\sqrt{23\times 23} \\ & =23 \end{align}
Hence the sum of two numbers is 23.

So, the correct answer is “Option B”.

Note: Since the values of $\left( a\times b \right)$ and $\left( {{a}^{2}}+{{b}^{2}} \right)$ are given in the question, don’t get confused while using the identity. By mistake, you might use the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
Using this formula, we will get the difference between two numbers which is not asked in the question. Hence, all the procedures will go the wrong way. So, be careful about what is asked and use the correct formula to get the answer.