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# The sum of the squares of two consecutive multiples of 7 is 1225. Find multiples.

Last updated date: 11th Aug 2024
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Hint: First assume the terms which are multiple of 7. Then, square the numbers and add them. After that equate them with the value 1225. Then simplify it and find the roots of the quadratic equation by factorization method. After that substitute the values to get the multiples.

Complete step-by-step solution:
Here, we are given that the sum of squares of two consecutive multiples of 7 is 1225.
Now, here we are talking about two consecutive multiples of 7 i.e. the difference between both numbers will be 7.
Let us suppose the first number which is a multiple of 7 be $7n$ and the second will be $7n + 7$.
Now, it is given that the sum of squares of $7n$ and $\left( {7n + 7} \right)$ is 1225. Hence, we can write the equation as
$\Rightarrow {\left( {7n} \right)^2} + {\left( {7n + 7} \right)^2} = 1225$
Using the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to expand the term ${\left( {7n + 7} \right)^2}$ in the above equation,
$\Rightarrow 49{n^2} + 49{n^2} + 98n + 49 = 1225$
Move all terms on one side and simplify it,
$\Rightarrow 98{n^2} + 98n - 1176 = 0$
Take 98 commons from the terms,
$\Rightarrow 98\left( {{n^2} + n - 12} \right) = 0$
Divide both sides by 98,
$\Rightarrow {n^2} - n - 12 = 0$
Split the coefficient of $n$ in such a way that their multiple will be -12.
$\Rightarrow {n^2} - 4n + 3n - 12 = 0$
Take $n$ common from the first two terms and 3 from the last two terms,
$\Rightarrow n\left( {n - 4} \right) + 3\left( {n - 4} \right) = 0$
Take $\left( {n - 4} \right)$ from both terms,
$\Rightarrow \left( {n - 4} \right)\left( {n + 3} \right) = 0$
Equate each term with 0,
$\Rightarrow n + 3 = 0$ and $n - 4 = 0$
Move the constant part on the right side,
$\Rightarrow n = - 3$ and $n = 4$
For $n = - 3$, the first multiple is,
$\Rightarrow 7n = 7 \times - 3$
Simplify the term,
$\therefore 7n = - 21$
For $n = - 3$, the second multiple is,
$\Rightarrow 7n - 7 = 7 \times - 3 - 7$
Simplify the term,
$\therefore 7n - 7 = - 28$
For $n = 4$, the first multiple is,
$\Rightarrow 7n = 7 \times 4$
Simplify the term,
$\therefore 7n = 28$
For $n = 4$, the second multiple is,
$\Rightarrow 7n - 7 = 7 \times 4 - 7$
Simplify the term,
$\therefore 7n - 7 = 21$

Hence, the multiples are either $- 28, - 21$ or $21,28$.

Note: A quadratic equation is a polynomial equation of degree 2. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution, or two imaginary solutions.
There are several methods you can use to solve a quadratic equation:
Factoring
Completing the Square