Question

find the two consecutive odd postive integers sum of whose squares is 290​

Answer 1

Basic Concept Used :-

Writing System of  Equation from Word Problem.

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.

Let's solve the problem now!!!

Solution :-

• Let the first odd number be x

☆ So,

• Other consecutive odd number be x + 2.

According to statement,

☆ Sum of the squares of two consecutive odd positive integer is 290.

$\rm :\longmapsto\: {x}^{2} + {(x + 2)}^{2} = 290$

$\rm :\longmapsto\: {x}^{2} + {x}^{2} + 4 + 4x = 290$

$\rm :\longmapsto\: 2{x}^{2} + 4 + 4x = 290$

$\rm :\longmapsto\: 2{x}^{2} + 4 + 4x - 290 = 0$

$\rm :\longmapsto\: 2{x}^{2} + 4x - 286 = 0$

$\rm :\longmapsto\: {x}^{2} + 2x - 143 = 0$

$\rm :\longmapsto\: {x}^{2} + 13x - 11x - 143 = 0$

$\rm :\longmapsto\:x(x + 13) - 11(x + 13) = 0$

$\rm :\longmapsto\:(x + 13)(x - 11) = 0$

$\rm :\implies\:x = 11 \: or \: x = - 13 \: \red{\{rejected \}}$

$\begin{gathered}\begin{gathered}\bf\: Hence \: numbers \: are-\begin{cases} &\sf{x = 11} \\ &\sf{x + 2 = 11 + 2 = 13} \end{cases}\end{gathered}\end{gathered}$