# The sum of the squares of two consecutive even numbers is 340. The product of the number is:

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Hint:Two consecutive numbers are numbers that have differences as one. We can write two consecutive even numbers as 2n and 2n+2. Find n using the given condition. Roots of quadratic equation $A{{x}^{2}}+Bx+C=0$ can be given by the formula, $\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$.

Here, we are given that the sum of squares of two consecutive even numbers is 340.

As, we know that consecutive numbers are numbers with the difference 1 i.e. 1,2,3,4.....type.

Now, here we are talking about two consecutive even numbers i.e. the difference between both numbers will be two. As any two consecutive numbers will have odd numbers in between them.

Now, we know that an even number is multiple of ‘2’ and can be given in terms of a variable as ‘2n’ which represents all even numbers for n = 1,2,3,4,..............

Hence, let us suppose the first even number be ‘2n’ and therefore second will be (2n+2) by the above mentioned rule.

Now, it is given that the sum of squares of 2n and (2n+2) is 340. Hence, we can write equation as

${{\left( 2n \right)}^{2}}+{{\left( 2n+2 \right)}^{2}}=340................\left( 1 \right)$

We have ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

For expansion of ${{\left( 2n+2 \right)}^{2}}$, we have to use the above relation in equation (1). Now, we get equation (1) as;

$\begin{align}

& 4{{n}^{2}}+{{\left( 2n \right)}^{2}}+{{\left( 2 \right)}^{2}}+2.2n.2=340 \\

& 4{{n}^{2}}+4{{n}^{2}}+4+8n=340 \\

& 8{{n}^{2}}+8n-336=0 \\

\end{align}$

Dividing whole equation by 8 to both sides, we get;

${{n}^{2}}+n-42=0...........\left( 2 \right)$

We have a quadratic formula to find roots of a given quadratic.

Let we have equation $A{{x}^{2}}+Bx+C=0$

We can give roots by relation

$roots=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}.................\left( 3 \right)$

Using equation (3) with equation (2), to get roots,

We have, ${{n}^{2}}+n-42=0$

Here A = 1, B = 1, C = -42 on comparing with $A{{x}^{2}}+Bx+C=0$

Hence, from the equation (3), we get

$\begin{align}

& roots=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times \left( -{{42}} \right)\times 1}}{2} \\

& roots=\dfrac{-1\pm \sqrt{1+168}}{2} \\

& roots=\dfrac{-1\pm \sqrt{169}}{2} \\

\end{align}$

We know the square root of 169 is 13.

Hence, above equation can be re-written as

$roots=\dfrac{-1\pm 13}{2}$

Roots can be given as $\dfrac{-1+13}{2}$and $\dfrac{-1-13}{2}$i.e. 6 and -7.

Now, we have n = 6 and n = -7 both.

As we have already supposed that 2n is an even number which means n is a positive integer. Hence, we can ignore n = -7

Therefore n = 6.

And numbers are (2n, 2n+2) which is given as $2\times 6,2\times 6+2\text{ or }\left( 12,14 \right)$.

Therefore, we will get 340 by sum of squaring of two even consecutive numbers 12 and 14.

Now, we have to find the product of these numbers as asked in the problem.

Hence, on multiplying them we get;

$product=14\times 12=168$

Hence, the answer is 168.

Note: One can go wrong by supposing two even consecutive numbers as (2n, 2n+1), (n, n+2) where (2n+1) represent odd numbers and (n, n+2) represents general two numbers, that may be even or odd. Hence, the first even number is 2n, we take the second even number to be 2n+2.

One can take two consecutive even numbers as (2n-2, 2n) or 2n+2, (2n+4) or (2n+8, 2n+10) etc, where all numbers are multiple of 2 and have a difference of 2 as well. Hence, supposing the above numbers, we get the same answer as in solution.

One can find the roots of quadratic ${{n}^{2}}+n-42=0$ by factorization method.

Now by splitting the middle coefficient by two numbers which gives the multiplication of -42 and summation of 1.

Here, we are given that the sum of squares of two consecutive even numbers is 340.

As, we know that consecutive numbers are numbers with the difference 1 i.e. 1,2,3,4.....type.

Now, here we are talking about two consecutive even numbers i.e. the difference between both numbers will be two. As any two consecutive numbers will have odd numbers in between them.

Now, we know that an even number is multiple of ‘2’ and can be given in terms of a variable as ‘2n’ which represents all even numbers for n = 1,2,3,4,..............

Hence, let us suppose the first even number be ‘2n’ and therefore second will be (2n+2) by the above mentioned rule.

Now, it is given that the sum of squares of 2n and (2n+2) is 340. Hence, we can write equation as

${{\left( 2n \right)}^{2}}+{{\left( 2n+2 \right)}^{2}}=340................\left( 1 \right)$

We have ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

For expansion of ${{\left( 2n+2 \right)}^{2}}$, we have to use the above relation in equation (1). Now, we get equation (1) as;

$\begin{align}

& 4{{n}^{2}}+{{\left( 2n \right)}^{2}}+{{\left( 2 \right)}^{2}}+2.2n.2=340 \\

& 4{{n}^{2}}+4{{n}^{2}}+4+8n=340 \\

& 8{{n}^{2}}+8n-336=0 \\

\end{align}$

Dividing whole equation by 8 to both sides, we get;

${{n}^{2}}+n-42=0...........\left( 2 \right)$

We have a quadratic formula to find roots of a given quadratic.

Let we have equation $A{{x}^{2}}+Bx+C=0$

We can give roots by relation

$roots=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}.................\left( 3 \right)$

Using equation (3) with equation (2), to get roots,

We have, ${{n}^{2}}+n-42=0$

Here A = 1, B = 1, C = -42 on comparing with $A{{x}^{2}}+Bx+C=0$

Hence, from the equation (3), we get

$\begin{align}

& roots=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times \left( -{{42}} \right)\times 1}}{2} \\

& roots=\dfrac{-1\pm \sqrt{1+168}}{2} \\

& roots=\dfrac{-1\pm \sqrt{169}}{2} \\

\end{align}$

We know the square root of 169 is 13.

Hence, above equation can be re-written as

$roots=\dfrac{-1\pm 13}{2}$

Roots can be given as $\dfrac{-1+13}{2}$and $\dfrac{-1-13}{2}$i.e. 6 and -7.

Now, we have n = 6 and n = -7 both.

As we have already supposed that 2n is an even number which means n is a positive integer. Hence, we can ignore n = -7

Therefore n = 6.

And numbers are (2n, 2n+2) which is given as $2\times 6,2\times 6+2\text{ or }\left( 12,14 \right)$.

Therefore, we will get 340 by sum of squaring of two even consecutive numbers 12 and 14.

Now, we have to find the product of these numbers as asked in the problem.

Hence, on multiplying them we get;

$product=14\times 12=168$

Hence, the answer is 168.

Note: One can go wrong by supposing two even consecutive numbers as (2n, 2n+1), (n, n+2) where (2n+1) represent odd numbers and (n, n+2) represents general two numbers, that may be even or odd. Hence, the first even number is 2n, we take the second even number to be 2n+2.

One can take two consecutive even numbers as (2n-2, 2n) or 2n+2, (2n+4) or (2n+8, 2n+10) etc, where all numbers are multiple of 2 and have a difference of 2 as well. Hence, supposing the above numbers, we get the same answer as in solution.

One can find the roots of quadratic ${{n}^{2}}+n-42=0$ by factorization method.

Now by splitting the middle coefficient by two numbers which gives the multiplication of -42 and summation of 1.

Last updated date: 21st Sep 2023

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