Question

The sum of two numbers is 16 and the sum of their squares is 146.The larger number is​

Answer 1

Answer:

The answer is given in the picture

Step-by-step explanation:

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Answer 2

${\large{\tt{Given:-}}}$

There are two numbers whose sum is 16 and the sum of their square is 146

${\large{\tt{Need\;to\;find:-}}}$

The larger  number

${\large{\tt{Answer:-}}}$

Let us assume the numbers a and b respectively

$\begin{gathered}\tt a+b=16\\\tt a=16-b (1)\end{gathered}$

$\begin{gathered}\tt a^2+b^2=146\\ \tt From\; 1 \\ \tt (16-b)^2+b^2=146\\ \tt 256-2(16)(b) +b^2+b^2=146\\ \tt 256-32b+b^2+b^2=146\\ \tt 2b^2 -32b=146-256\\ \tt 2b^2-32b+110 =0\end{gathered}$

$\begin{gathered} \tt \dfrac{2b^2-32b+110}{2}=0\\ \tt b^2-16b+55=0\\ \tt b^2 - (11b + 5b) + 55 = 0\\\tt (b-11)(b-5)=0\\ \tt b=11\;\;\&\;\;5\end{gathered}$

Taking b as 5 will give

$\begin{gathered}\tt a=16-5\\ \tt a=11\end{gathered}$

Taking b as 11 will give

$\begin{gathered}\tt a=16-11\\ \tt a=5\end{gathered}$

11 is the larger number