Question

vo consecutive positive integers, sum ofwhose squares is 613.
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Answer 1

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Answer 2

$\huge \purple{ \underline{ \boxed{ \red{ \mathbb{ANSWER : }}}}}$

$\red{ \textsf{The two consecutive positive integers sum of whose squares is 613 are\orange{ 17 and 18.}}}$

$\huge \purple{ \underline{ \boxed{ \red{ \mathbb{EXPLANATION : }}}}}$

$\large\green{ \underline{ \underline{ \mathbb{GIVEN : }}}}$

$\orange{ \textsf{Sum of squares of two positive integers is 613 .}}$

$\large\green{ \underline{ \underline{ \mathbb{FORMULA \: USED : }}}}$

$\orange{ \textsf{( a + b ) ² = a² + b² + 2ab}}$

$\large\green{ \underline{ \underline{ \mathbb{SOLUTION : }}}}$

$\red{ \textsf{Let x be the first number and ( x + 1 ) be the second number . }}$

$\red{ \textsf{Therefore}}$

$\red{ \longmapsto \mathsf{ {x}^{2} + {(x + 1)}^{2} = 613}}$

$\red{ \longmapsto \mathsf{ {x}^{2} + {x}^{2} + 1 + 2x = 613}}$

$\red{ \longmapsto \mathsf{ {2x}^{2} + 2x + 1 = 613}}$

$\red{ \longmapsto \mathsf{ {2x}^{2} + 2x = 613 - 1}}$

$\red{ \longmapsto \mathsf{ {2x}^{2} + 2x = 612}}$

$\red{ \longmapsto \mathsf{ {2x}^{2} + 2x - 612 = 0}}$

$\red{ \longmapsto \mathsf{ 2({x}^{2} + x - 306) = 0}}$

$\red{ \longmapsto \mathsf{ {x}^{2} + x - 306 = 0}}$

$\red{ \longmapsto \mathsf{ {x}^{2} +18 x - 17x - 306 = 0}}$

$\red{ \longmapsto \mathsf{ x( x+18 ) - 17(x + 18) = 0}}$

$\red{ \longmapsto \mathsf{ ( x - 17)( x+18 )= 0}}$

$\red{ \begin{array}{l | l} \longmapsto \mathsf{ x - 17 = 0 }& \mathsf{x+18= 0} \\ \longmapsto \mathsf{ x = 17 }& \mathsf{x = - 18} \end{array}}$

$\red{ \textsf{Therefore ,}}$

$\red{ \textsf{First number = 17 or -18}}$

$\red{ \textsf{Second number = 18 or -17}}$

$\large \green{ \underline{ \underline{ \mathbb{KNOW \: MORE : }}}}$

$\orange{\mathbb{INTEGERS : }}$

An integer is a whole number not a fractional number that can be positive , negative , or zero .

$\orange{ \mathbb{SQUARE \: NUMBER :}}$

A square number or perfect square is an integer that is the square of an integer ; in other words , it is the product of some integer with itself .