(m) If $$ x-\frac{1}{x}=7 $$, detarmine the value of: (1) $$ x^{2}+\frac{1}{x^{2}} $$

Question

UpStudy AI · Accepted Answer

To determine the value of $$ x^{2} + \frac{1}{x^{2}} $$ given that $$ x - \frac{1}{x} = 7 $$, follow these steps:

1. **Square Both Sides of the Given Equation**

Start by squaring both sides of the equation $$ x - \frac{1}{x} = 7 $$:

$$
   \left( x - \frac{1}{x} \right)^2 = 7^2
   $$

$$
   x^2 - 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 49
   $$

Simplify the middle term:

$$
   x^2 - 2 + \frac{1}{x^2} = 49
   $$

2. **Isolate $$ x^{2} + \frac{1}{x^{2}} $$**

Add 2 to both sides of the equation to solve for $$ x^{2} + \frac{1}{x^{2}} $$:

$$
   x^2 + \frac{1}{x^2} = 49 + 2
   $$

$$
   x^2 + \frac{1}{x^2} = 51
   $$

**Answer:** The value of $$ x^{2} + \frac{1}{x^{2}} $$ is **51**.
$$ x^{2} + \frac{1}{x^{2}} = 51 $$

(m) If \( x-\frac{1}{x}=7 \), detarmine the value of: (1) \( x^{2}+\frac{1}{x^{2}} \)