Question

Sum of squares of two consecutive even numbers is 580. Find the number by writing a suitable quadratic equation.

Answer 1

Hint: In this question, we use the concept of even consecutive numbers. Even numbers are numbers that end with 0, 2, 4, 6 or 8. Two consecutive even numbers can be written as 2n and 2n+2, where n is any integer. The examples of consecutive even numbers are ….,-4-2,0, 2, 4, 6, 8, 10, 12, ....

Complete step-by-step answer:

Let 2n and 2n+2 be the required consecutive even numbers.
Given that the sum of squares of two consecutive even numbers is 580.
Therefore,
$
  {\left( {2n} \right)^2} + {\left( {2n + 2} \right)^2} = 580 \\
   \Rightarrow 4{n^2} + 4{n^2} + 8n + 4 = 580 \\
   \Rightarrow 8{n^2} + 8n - 576 = 0..........\left( 1 \right) \\
 $
Take common 8 from (1) equation
$
   \Rightarrow 8\left( {{n^2} + n - 72} \right) = 0 \\
   \Rightarrow {n^2} + n - 72 = 0 \\
$
Quadratic equation is ${n^2} + n - 72 = 0$
Now factories by splitting the middle term.
$
   \Rightarrow \left( {n + 9} \right)\left( {n - 8} \right) = 0 \\
   \Rightarrow n = 8, - 9 \\
 $
Case 1: If n=8
Therefore, the required numbers are,
$
  2n = 2 \times 8 = 16 \\
  2n + 2 = 2 \times 8 + 2 = 18 \\
$
Case 2: If n=-9
Therefore, the required numbers are,
$
  2n = 2 \times \left( { - 9} \right) = - 18 \\
  2n + 2 = 2 \times \left( { - 9} \right) + 2 = - 16 \\
$
So, the required numbers are 16 and 18 or -18 and -16.

A suitable quadratic equation is ${n^2} + n - 72 = 0$.

Note: Whenever we face such types of problems we use some important points. First we assume the consecutive even numbers and make an equation according to the given question then after solving the quadratic equation we can get the required answer.