Question

Sum of two consecutive square of number are 481. Find the sum of the two number

Answer 1

let first number = X

let second number = X+1

square of the two number

(X)^2 + (X+1)^2 =481

x^2 + x^2+1+2x =481

2x^2 +2x +1-481

2(X2+x-240) =0

X2 +X -240

X2 +16x -15x -240

X(X+16) -15(X+16)

x-15) (X+16)

X=15

X=-16

Answer 2

$\large\bf{\underline{\underline{Question:−}}}$

The sum of the squares of two consecutive positive integers is 481. Find the integers.

$\large\bf{\underline{\underline{Answer:−}}}$

$\color{red}\longrightarrow{}$$\rm{x² + (x+1)^2 = 481}$

$\color{red}\longrightarrow{}$$\rm{x^2 + x^2 + 2x + 1 = 481}$

$\color{red}\longrightarrow{}$$\rm{2x^2 + 2x + 1 - 481 = 0}$

$\color{red}\longrightarrow{}$$\rm{2x^2 + 2x - 480 = 0}$

$\underline\bold\color{purple}{Simplify,}$

$\sf{Divide\: by\: 2}$

$\rm{x^2 + x - 240 = 0}$

$\sf{Factors \:to}$

$\rm{(x+16)(x-15) = 0}$

$\sf{positive \:solution}$

x=15 & 16 are the integers

$\tt{See \:if\: that\: works}$

$\color{teal}\underline{\underline{ \tt{{ \boxed {15^2 + 16^2 = 225 + 256 = 481} }}}}$

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Thankyou :)