Show that if $$ x^{2}-2 x

Question

Show that if $$ x^{2}-2 x<5 $$ then $$ x-1< $$

UpStudy AI · Accepted Answer

It seems like you're looking to establish an inequality involving $$ x - 1 $$ based on the condition $$ x^{2} - 2x < 5 $$. Let's proceed step by step to show that:

$$
\text{If } x^{2} - 2x < 5 \text{, then } x - 1 < \sqrt{6}.
$$

### Step 1: Start with the Given Inequality
You have the inequality:

$$
x^{2} - 2x < 5
$$

### Step 2: Complete the Square
To make the left side a perfect square, follow these steps:

1. **Find the coefficient of $$ x $$**: Here, the coefficient is $$-2$$.
2. **Take half of it and square it**:

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

3. **Add and subtract this number inside the inequality**:

$$
x^{2} - 2x + 1 - 1 < 5
$$

4. **Rewrite the left side as a squared term**:

$$
(x - 1)^2 - 1 < 5
$$

### Step 3: Simplify the Inequality
Add 1 to both sides to isolate the squared term:

$$
(x - 1)^2 < 6
$$

### Step 4: Take the Square Root of Both Sides
To solve for $$ x - 1 $$, take the square root of both sides, remembering to consider both the positive and negative roots:

$$
|x - 1| < \sqrt{6}
$$

This inequality tells us that the distance between $$ x $$ and 1 is less than $$ \sqrt{6} $$.

### Step 5: Derive the Final Inequality
From the absolute value inequality, we have:

$$
- \sqrt{6} < x - 1 < \sqrt{6}
$$

Focusing on the upper bound:

$$
x - 1 < \sqrt{6}
$$

### Conclusion
Therefore, starting from the condition $$ x^{2} - 2x < 5 $$, we've shown through completing the square and simplifying that:

$$
x - 1 < \sqrt{6}
$$

This concludes the proof.
If $$ x^{2} - 2x < 5 $$, then $$ x - 1 < \sqrt{6} $$.

Show that if
then

Upstudy AI Solution

Answer

Solution

Step 1: Start with the Given Inequality

Step 2: Complete the Square

Step 3: Simplify the Inequality

Step 4: Take the Square Root of Both Sides

Step 5: Derive the Final Inequality

Conclusion

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