Question

The sum of a squares of two consecutive position even number is 52. find numbers.

Answer 1

Let the even number be x.
The other number will be (x+2)
The sum of their squares is given 52.
${x}^{2} + {(x + 2)}^{2} = 52$
${x}^{2} + {x}^{2} + 4x + 4 = 52$
$2 {x}^{2} + 4x + 4 - 52 = 0$
$2 {x}^{2} + 4x - 48 = 0$
${x}^{2} + 2x - 24 = 0$

We now have a quadratic equation.
We have to find two numbers such that their subtraction will be the middle term and whose multiplication will be the multiplication of coefficients of first and last term.
Therefore,
${x}^{2} + 6x - 4x - 24 = 0$
$x(x + 6) - 4(x + 6) = 0$
$(x - 4)(x + 6) = 0$
Now either
$(x - 4) = 0$
or
$(x + 6) = 0$
Therefore,
$x = 4 \: or \: - 6$
But we need a positive number,
Therefore,
$x = 4$
We know that second number is
$x + 2$
$4 + 2$
$6$
Therefore, the two numbers are 4 and 6

Answer 2

Let the first number be x²

Let the second number which is consecutive even number be x² + 2

Given, their sum is 52


According to the given condition,

x² + (x² + 2) = 52

➾ x² + x² + 4x + 4 = 52

➾ x² + x² + 4x = 52 - 4

➾ 2x² + 4x = 48

➾ 2x² + 4x - 48 = 0

➾ x² + 2x - 24 = 0

➾ (x + 6) (x - 4) = 0


x + 6 = 0 => x = -6 => This is not a positive number

x - 4 = 0 => x = 4 => This is a positive number


1 consecutive even number cannot be negative.

∴ The 2 numbers are $\green{\boxed{\green{\boxed{\red{\textsf{4\:and\:6}}}}}}$