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Find two consecutive positive integers, the sum of whose squares is 313.

Last updated date: 27th Mar 2023
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Answer
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Hint: Let the smaller number be $a$,then the second number will be $a+1$. Square these and add them and then equate the sum to 313. Then solve for $a$.

Complete step-by-step answer:
Let the smaller number be $a$. Now since the numbers are consecutive, the second number will be $a+1$.
We are given that the sum of their squares is equal to 313.
${{a}^{2}}+{{\left( a+1 \right)}^{2}}=313$ …(1)
We know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. We will use this to simplify ${{\left( a+1 \right)}^{2}}$
${{\left( a+1 \right)}^{2}}={{a}^{2}}+1+2a$
We will substitute this in equation (1)
${{a}^{2}}+{{a}^{2}}+1+2a=313$
$2{{a}^{2}}+2a+1=313$
$2{{a}^{2}}+2a-312=0$
Divide this by 2, we will get the following:
${{a}^{2}}+a-156=0$
This is a quadratic equation in $a$.
We know that $x=\dfrac{b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$where $a{{x}^{2}}+bx+c=0$ is a quadratic equation in $x$ .
Using this, we get the following:
$a=\dfrac{-1\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -156 \right)}}{2\times 1}$
$a=\dfrac{-1\pm \sqrt{1+624}}{2}$
$a=\dfrac{-1\pm \sqrt{625}}{2}$
$a=\dfrac{-1\pm 25}{2}$
$a=\dfrac{-26}{2},\dfrac{24}{2}$
$a=-13,12$
Since, in the question we need positive integers so we will select $a=12$
So $a+1=13$
Hence, the required consecutive positive integers are $12$ and $13$.

Note: You can check whether your answer is correct or not by substituting $a=12$ and $a+1=13$ in the given equation ${{a}^{2}}+{{\left( a+1 \right)}^{2}}=313$. We get the following:
${{12}^{2}}+{{13}^{2}}=144+169=313$. So our answer is correct.