Question: show that the rootis of the eqaution $x$ squared $- 3x + 2 - c$ squared $= 0$ are real
1. **State the problem:**
We need to show that the roots of the equation $$x^2 - 3x + 2 - c^2 = 0$$ are real.
2. **Recall the quadratic formula:**
For a quadratic equation $$ax^2 + bx + c = 0$$, the roots are given by:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The roots are real if and only if the discriminant $$\Delta = b^2 - 4ac \geq 0$$.
3. **Identify coefficients:**
In our equation, $$a = 1$$, $$b = -3$$, and $$c = 2 - c^2$$ (note: this is the constant term, not the variable $$c$$).
4. **Calculate the discriminant:**
$$\Delta = (-3)^2 - 4 \times 1 \times (2 - c^2) = 9 - 4(2 - c^2)$$
5. **Simplify the discriminant:**
$$\Delta = 9 - 8 + 4c^2 = 1 + 4c^2$$
6. **Analyze the discriminant:**
Since $$c^2 \geq 0$$ for all real $$c$$, then $$4c^2 \geq 0$$.
Therefore, $$\Delta = 1 + 4c^2 \geq 1 > 0$$.
7. **Conclusion:**
Because $$\Delta > 0$$ for all real $$c$$, the roots of the equation $$x^2 - 3x + 2 - c^2 = 0$$ are always real.
**Final answer:** The roots are real for all real values of $$c$$.