The multiplication of two numbers is 120 and the sum of their square is 289. Find their sum.
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The multiplication of two numbers is 120 and the sum of their square is 289. Find their sum.

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Solution :

 Let the numbers are a and $b$.

$\therefore$ According to question,

and $a^{2}+b^{2}=289$

We know that, $$ \begin{aligned}(a+b)^{2} &=a^{2}+b^{2}+2 a b \\ &=289+2 \times 120 \\ &=289+240=529 \end{aligned} $$ $\therefore \quad a+b=\sqrt{529}=23 ;$ Ans.

$$

\begin{aligned}

(a+b)^{2} &=a^{2}+b^{2}+2 a b \\

&=289+2 \times 120 \\

&=289+240=529 \\

\therefore a+b=& \sqrt{529}=23 ; \text { Ans. }

\end{aligned}

$$



Trick :

$$

\begin{aligned}

a+b &=\sqrt{289+2 \times 120} \\

&=\sqrt{529}=23: \text { Ans. }

\end{aligned}

$$

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