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Find the consecutive even integers whose squares have the sum 340.

Last updated date: 20th Jun 2024
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Hint: In this question, we will be letting the two consecutives even integers as variables. Then form a quadratic equation with the given data. Further simplify it by splitting and grouping the common terms to get the required values.

Let us consider the two consecutives even integers as $2x$ and $2x + 2$ where $x$ is a positive integer.
Given that these integers sum of squares is 34. So, we have
$\Rightarrow {\left( {2x} \right)^2} + {\left( {2x + 2} \right)^2} = {\left( {340} \right)^2}$
We know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. By using this formula, we have
$\Rightarrow 4{x^2} + 4{x^2} + 4 + 8x = 340$
And hence on further simplification, we have
$\Rightarrow 8{x^2} + 8x + 400 - 4 = 0 \\ \Rightarrow 8{x^2} + 8x - 336 = 0 \\$
Now on taking 8 common from the whole equation, we have
$\Rightarrow 8\left( {{x^2} + x - 42} \right) = 0 \\ \Rightarrow {x^2} + x - 42 = 0 \\$
By splitting and grouping the common terms, we have
$\Rightarrow {x^2} + 7x - 6x - 42 = 0 \\ \Rightarrow x\left( {x + 7} \right) - 6\left( {x + 7} \right) = 0 \\ \Rightarrow \left( {x + 7} \right)\left( {x - 6} \right) = 0 \\ \therefore x = 6, - 7 \\$
Since, $x$ is a positive integer we have $x = 6$
Hence, the two consecutives even integers are $2x = 2 \times 6 = 12$ and $2x + 2 = 2 \times 6 + 2 = 14$.
Thus, the required values are 12, 14.

Note: Here the formed equation ${x^2} + x - 42 = 0$ is a quadratic equation. A quadratic equation is an equation in one variable whose degree is 2 and has two roots or zeros. A number which is exactly divisible by 2 without leaving any remainder is called an even number.