Question

One positive integer is 2 less than twice another. The sum of their squares is 628. Find the integers.

Answer 1

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let one of the integers be x,

other one = 2x - 2

now, according to question :

• ${x}^{2} + (2x - 2) {}^{2} = 628$

• ${x}^{2} + 4 {x}^{2} + 4 - 8 x = 628$

• $5 {x}^{2} - 8x = 628 - 4$

• $5 {x}^{2} - 8x = 624$

• $5 {x}^{2} - 8x - 624 = 0$

• $5 {x}^{2} - 60 x+ 52 x- 624 = 0$

• $5x(x - 12) + 52(x - 12) = 0$

• $(5x + 52)(x - 12) = 0$

now, since the Integers are positive so the value obtained from (5x + 52) = 0 can't hold true.

so, x - 12 = 0

• $x = 12$

the first number is :

• $x = 12$

second number is :

• $2x - 2$

• $(2 \times 12) - 2$

• $24 - 2$

• $22$