4.1 Discuss the nature of the roots of $$ x^{2}-2 x-6=0 $$, without solving the 4.2 The roots of a quadratic equation are $$ x=\frac{-2 \pm \sqrt{13-2 k}}{3} $$. 1.4.4.1 For which values of $$ k $$ will the roots be real?

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## **4.1. Discuss the Nature of the Roots of $$ x^{2} - 2x - 6 = 0 $$ Without Solving**

To determine the nature of the roots of a quadratic equation without actually solving it, we use the **discriminant**. The discriminant provides crucial information about the roots based on its value.

### **Quadratic Equation Standard Form**
A general quadratic equation is given by:
$$ ax^{2} + bx + c = 0 $$

### **Discriminant (D)**
The discriminant is calculated as:
$$ D = b^{2} - 4ac $$

### **Interpreting the Discriminant**

1. **If $$ D > 0 $$:**
   - The equation has **two distinct real roots**.
   
2. **If $$ D = 0 $$:**
   - The equation has **exactly one real root** (a repeated root).
   
3. **If $$ D < 0 $$:**
   - The equation has **two complex (imaginary) conjugate roots**.

### **Applying to $$ x^{2} - 2x - 6 = 0 $$**
Identify $$ a $$, $$ b $$, and $$ c $$:
- $$ a = 1 $$
- $$ b = -2 $$
- $$ c = -6 $$

#### **Calculate the Discriminant:**
$$
D = (-2)^2 - 4(1)(-6) = 4 + 24 = 28
$$

#### **Interpretation:**
- Since $$ D = 28 > 0 $$, the quadratic equation has **two distinct real roots**.

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## **4.2. The Roots of a Quadratic Equation are $$ x = \frac{-2 \pm \sqrt{13 - 2k}}{3} $$**

### **1.4.4.1. For Which Values of $$ k $$ Will the Roots Be Real?**

The given roots are expressed in the form:
$$
x = \frac{-2 \pm \sqrt{13 - 2k}}{3}
$$

For the roots to be **real**, the expression under the square root (the **radicand**) must be **non-negative**.

### **Determine When the Radicand is Non-Negative:**
$$
13 - 2k \geq 0
$$

### **Solving the Inequality:**
$$
13 - 2k \geq 0 \
-2k \geq -13 \quad \text{(Subtracting 13 from both sides)} \
2k \leq 13 \quad \text{(Multiplying both sides by -1, which reverses the inequality)} \
k \leq \frac{13}{2} \
k \leq 6.5
$$

### **Conclusion:**
- **All real numbers $$ k $$ such that $$ k \leq 6.5 $$** will ensure that the roots of the quadratic equation are **real**.
- If $$ k > 6.5 $$, the radicand becomes negative, resulting in **complex (imaginary) roots**.

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**Summary:**
- **For $$ x^{2} - 2x - 6 = 0 $$:** There are **two distinct real roots**.
- **For the roots $$ x = \frac{-2 \pm \sqrt{13 - 2k}}{3} $$:** The roots are **real** if and only if $$ k \leq 6.5 $$.
- The equation $$ x^{2} - 2x - 6 = 0 $$ has two distinct real roots. - The roots $$ x = \frac{-2 \pm \sqrt{13 - 2k}}{3} $$ are real when $$ k \leq 6.5 $$.

4.1 Discuss the nature of the roots of , without solving the
4.2 The roots of a quadratic equation are .
1.4.4.1 For which values of will the roots be real?

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4.1 Discuss the nature of the roots of , without solving the 4.2 The roots of a quadratic equation are . 1.4.4.1 For which values of will the roots be real?