Question

Two consecutive positive integers, sum of whose squares is 365 are
\[
  \left( a \right)13,14 \\
  \left( b \right)14,15 \\
  \left( c \right)12,13 \\
  \left( d \right)11,12 \\
\]

Answer 1

- Hint-In this question, we used the concept quadratic equation and found the root of the quadratic equation by using the sridharacharya formula. If quadratic equation $a{x^2} + bx + c = 0$ so the roots of this quadratic equation by sridharacharya formula is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .In this question we have to assume two consecutive positive integers are x and (x+1) where x is non negative integer.


Complete step-by-step solution -

Let two consecutive positive integer be x and x+1.
Now, the sum of squares of two consecutive positive integers is 365 so we can write it in mathematical form.
\[ \Rightarrow {\left( x \right)^2} + {\left( {x + 1} \right)^2} = 365\]
We use algebraic identity, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
\[
   \Rightarrow {\left( x \right)^2} + {\left( x \right)^2} + 2 \times x \times 1 + {\left( 1 \right)^2} = 365 \\
   \Rightarrow 2{x^2} + 2x + 1 = 365 \\
   \Rightarrow 2{x^2} + 2x - 364 = 0 \\
   \Rightarrow {x^2} + x - 182 = 0 \\
 \]
Now, we can see the quadratic equation in x so we use the sridharacharya formula to find roots. If quadratic equation $a{x^2} + bx + c = 0$ so the roots of this quadratic equation by sridharacharya formula is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
$
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 1 \times \left( { - 182} \right)} }}{{2 \times 1}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 728} }}{{2 \times 1}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {729} }}{2} \\
$
We use square root of $\sqrt {729} = 27$
$ \Rightarrow x = \dfrac{{ - 1 \pm 27}}{2}$
If we take negative value,
\[
   \Rightarrow x = \dfrac{{ - 1 - 27}}{2} = \dfrac{{ - 28}}{2} \\
   \Rightarrow x = - 14 \\
\]
We assume only positive values of x so we have to eliminate negative values.
Now, we take positive value
$
   \Rightarrow x = \dfrac{{ - 1 + 27}}{2} \\
   \Rightarrow x = \dfrac{{26}}{2} \\
   \Rightarrow x = 13 \\
 $
Two consecutive positive integers are x=13 and x+1=14.
So, the correct option is (a).
Note-In such types of problems we generally face problems with roots of quadratic equations. So, we can use another method to find roots of the quadratic equation that is splitting the middle term method. Let’s take a quadratic equation ${x^2} + 3x + 2 = 0$ so we split middle term as product of coefficient of highest degree (1) and constant (2) is 2 then their sum is equal to coefficient of second higher degree, 2+1=3. So, we can write ${x^2} + \left( {2 + 1} \right)x + 2 = 0 \Rightarrow {x^2} + 2x + x + 2 = 0$ then we take common, $x\left( {x + 2} \right) + 1 \times \left( {x + 2} \right) \Rightarrow \left( {x + 2} \right)\left( {x + 1} \right)$ .