Question

Find two rational numbers which differ by 3 and the sum of whose squares is 117.
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Answer 1

$\large\underline{\sf{Solution-}}$

↝Let assume that

• First rational number be x.

↝Since, two rational numbers differ by 3

So,

• Second rational number be x + 3.

According to statement,

↝Sum of the squares of two rational numbers is 117.

$\rm :\longmapsto\: {x}^{2} + {(x + 3)}^{2} = 117$

$\rm :\longmapsto\: {x}^{2} + {x}^{2} + 9 + 6x = 117$

$\rm :\longmapsto\: 2{x}^{2} + 9 + 6x - 117 = 0$

$\rm :\longmapsto\: 2{x}^{2} + 6x - 108 = 0$

↝ On dividing by 2, we get

$\rm :\longmapsto\: {x}^{2} + 3x - 54 = 0$

$\rm :\longmapsto\: {x}^{2} + 9x - 6x - 54 = 0$

$\rm :\longmapsto\:x(x + 9) - 6(x + 9) = 0$

$\rm :\longmapsto\:(x + 9)(x - 6) = 0$

$\bf\implies \:x = 6 \: \: \: or \: \: \: x = - 9$

Hence,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf x + 3 \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 6 & \sf 9 \\ \\ \sf - 9 & \sf - 6 \end{array}} \\ \end{gathered}$

Thus,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Numbers \: are \: -\begin{cases} &\sf{ \: \: 6 \: \: and \: \: 9} \\ &\sf{ \: \: \: \: \: \: or}\\ &\sf{ - 9 \: and \: - 6} \end{cases}\end{gathered}\end{gathered}$

Basic Concept Used :-

Writing Systems 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.

Answer 2

*Question:—

Find two rational numbers which differ by 3 and the sum of whose squares is 117.

*Answer:—

→ Let the 2 numbers be a,b

Given that :—

→ a − b = 3 ______________________(1)

→ a²+b²=117______________________(2)

Eq(1) :— a−b=3

Squaring on both sides:—

→ (a − b)² = 9

→ a² + b² − 2ab = 9

Substituting the value of a² + b² from eq (2) in eq (1) :—

→ 117 − 2ab = 9

→ - 2ab = 9 - 117

→ - 2ab = - 108

→ 2ab = 108

→ ab = 54__________________________(3)

→ b = 54/a

Substituting this value of b in eq (1),we get:—

→ a − 54/a = 3

→ a² − 54 = 3a

→ a² − 3a − 54 = 0

Solving this quadratic equation, we get :—

→ a=9,−6

But,

→ - 6 is not the solution because it is given that they are natural numbers.

So,

→ a = 9

Therefore:—

→ b = 9 − 3 = 6

→ b = 6

Hence our answer is 6.

$@vikramjeeth$