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$\left( i \right)$ Find three successive even natural numbers, the sum of whose squares is $308$.$\left( ii \right)$ Find three odd consecutive integers, the sum of whose squares is $83$.

Last updated date: 21st Jul 2024
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Hint: In order to solve the given problems, we must be considering a variable value and then assign the values to the variables accordingly for even numbers as they are successive. The difference between each of the successive even natural numbers would be $2$. And in the second case, the difference between odd consecutive integers would be $2$. We must form the equation which sums up the numbers and the squaring them should be equated to the given values. The obtained numbers would be the required solution.

Complete step-by-step solution:
Now let us have a brief regarding linear equations. Now let us learn about linear equations. A linear equation can be expressed in the form any number of variables as required. As the number of the variables increase, the name of the equation simply denotes it. The general equation of a linear equation in single variable is
$ax+b=0$
We can find the linear equation in three major ways. They are: point-slope form, standard form and slope-intercept form.
Now let us consider the first case.
Let one of the numbers be$2x$ as it is even.
Since the numbers must be successive even natural numbers, the numbers would be
$2x,2x+2,2x+4$
It is given that sum of squares of numbers is equal to $308$
Now writing it accordingly, we get
${{\left( 2x \right)}^{2}}+{{\left( 2x+2 \right)}^{2}}+{{\left( 2x+4 \right)}^{2}}=308$
Upon solving this equation, we get
\begin{align} & \Rightarrow {{\left( 2x \right)}^{2}}+{{\left( 2x+2 \right)}^{2}}+{{\left( 2x+4 \right)}^{2}}=308 \\ & \Rightarrow 4{{x}^{2}}+\left( 4{{x}^{2}}+8x+4 \right)+\left( 4{{x}^{2}}+16x+16 \right)-308=0 \\ & \Rightarrow 12{{x}^{2}}+24x-288=0 \\ \end{align}
Before we factorize, let us take out the common factor value and simplify it. We get,
\begin{align} & \Rightarrow 12{{x}^{2}}+24x-288=0 \\ & \Rightarrow 12\left( {{x}^{2}}+2x-24 \right)=0 \\ & \Rightarrow {{x}^{2}}+2x-24=0 \\ \end{align}
Upon factoring the equation, we obtain,
\begin{align} & \Rightarrow {{x}^{2}}+2x-24=0 \\ & \Rightarrow {{x}^{2}}+6x-4x-24=0 \\ & \Rightarrow x(x+6)-4(x+6)=0 \\ & \Rightarrow (x+6)(x-4)=0 \\ \end{align}
$\therefore x=-6,4$
Since we are considering only natural numbers, we will be considering $x=4$.
Now, the required numbers are:
\begin{align} & 2x=2\left( 4 \right)=8 \\ & 2x+2=2\left( 4 \right)+2=10 \\ & 2x+4=2\left( 4 \right)+4=12 \\ \end{align}
$\therefore$ The three successive even natural numbers, the sum of whose squares is $308$ is $8,10,12$.
Now let's consider the second case.
Let one of the odd numbers be $x$.
The other consecutive odd numbers would be $x+2,x+4$
From the question, the squares of these numbers would be equal to $83$.
So,
${{x}^{2}}+{{\left( x+2 \right)}^{2}}+{{\left( x+4 \right)}^{2}}=83$
Upon solving this, we get
${{x}^{2}}+{{x}^{2}}+4x+4+{{x}^{2}}+8x+16-83=0$
$\Rightarrow 3{{x}^{2}}+12x-63=0$
Upon taking out the common factor value from the equation, we get
\begin{align} & \Rightarrow 3{{x}^{2}}+12x-63=0 \\ & \Rightarrow {{x}^{2}}+4x-21=0 \\ \end{align}
Upon factoring it, we get
\begin{align} & \Rightarrow {{x}^{2}}+4x-21=0 \\ & \Rightarrow {{x}^{2}}+7x-3x-21=0 \\ & \Rightarrow x(x+7)-3(x+7)=0 \\ & \Rightarrow (x+7)(x-3)=0 \\ \end{align}
$\therefore x=-7,3$
Here, we will be considering both the cases, as we are given to find the integers.
Hence the numbers would be-
When $x=-7$,
\begin{align} & x=-7 \\ & x+2=-7+2=-5 \\ & x+4=-7+4=-3 \\ \end{align}
When $x=3$,
\begin{align} & x=3 \\ & x+2=3+2=5 \\ & x+4=3+4=7 \\ \end{align}
$\therefore$ The three successive odd natural numbers, the sum of whose squares is $83$ is $3,5,7$.

Note: We must consider such values of $x$, so that the chosen value satisfies with the given condition. If we are given the condition of natural numbers, then we are not supposed to choose negative values for $x$. We must note this point as it is applicable to all such problems.