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# If $2x+y=14$ and $xy=6$, find the value of $4{{x}^{2}}+{{y}^{2}}$.  Verified
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Hint: Here, we have to square the first equation $2x+y=14$ and then apply the second equation $xy=6$ into the first one.

Here, we are given with two equations:
$2x+y=14\text{ }.....\text{ (1)}$
$xy=6\text{ }.....\text{ (2)}$
With the help of these two equations we have to find the value of $4{{x}^{2}}+{{y}^{2}}$.
First, let us consider the equation (1).
By squaring equation (1) on both the sides we get,
${{(2x+y)}^{2}}={{14}^{2}}\text{ }....\text{ (3)}$
From equation (3) we can say that its LHS is of the form ${{(a+b)}^{2}}$whose expansion we are familiar with. i.e. the expansion for ${{(a+b)}^{2}}$is given as:
${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
In the LHS of equation (3), has the form ${{(a+b)}^{2}}$ where $a=2x$, and $b=y$.
Now by applying the above formula in equation (3) we get:
\begin{align} & {{(2x)}^{2}}+2\times 2x\times y+{{y}^{2}}=196 \\ & 4{{x}^{2}}+4xy+{{y}^{2}}=196 \\ \end{align}
By rearranging the equation we obtain:
$4{{x}^{2}}+{{y}^{2}}+4xy=196\text{ }...\text{ (4)}$
In the LHS of equation (4) we have $4xy$, but we know that $xy=6$. i.e.
In the next step we have to apply equation (2) in equation (4). Hence, we get the equation:
$4{{x}^{2}}+{{y}^{2}}+4\times 6=196$ i.e. by putting $xy=6\text{ }$
$4{{x}^{2}}+{{y}^{2}}+24=196$
In the next step, take 24 to the right side then 24 becomes -24, i.e. take variables to one side and constants to the other side. When the side changes, the sign also changes. i.e. we get the equation:
\begin{align} & 4{{x}^{2}}+{{y}^{2}}=196-24 \\ & 4{{x}^{2}}+{{y}^{2}}=172 \\ \end{align}
Hence, the value of $4{{x}^{2}}+{{y}^{2}}=172$

Note: Here don’t try to solve the equation (1) by putting $y=\dfrac{6}{x}$, it will become more complicated and also there is a large probability that the answer would be wrong. It is better to solve this by squaring the first equation $2x+y=14$.