# If $2x+y=14$ and $xy=6$, find the value of $4{{x}^{2}}+{{y}^{2}}$.

Last updated date: 22nd Mar 2023

•

Total views: 304.2k

•

Views today: 7.83k

Answer

Verified

304.2k+ views

Hint: Here, we have to square the first equation $2x+y=14$ and then apply the second equation $xy=6$ into the first one.

Complete step-by-step answer:

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 $2x+y=14$.

Complete step-by-step answer:

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 $2x+y=14$.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE