𝑋2−9𝑋 +14= 0
 5𝑋2−6𝑋 +1= 0
 6𝑋2−24= 0
 5𝑋2+10= 0
6𝑋2−9𝑋 = 0

Let's solve each quadratic equation step by step:

### 1. \( X^2 - 9X + 14 = 0 \)
This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -9 \), and \( c = 14 \).

We can solve this using the **quadratic formula**:
\[
X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substitute \( a = 1 \), \( b = -9 \), and \( c = 14 \):
\[
X = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)} = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2}
\]
\[
X = \frac{9 \pm 5}{2}
\]
So, the two solutions are:
\[
X = \frac{9 + 5}{2} = 7 \quad \text{or} \quad X = \frac{9 - 5}{2} = 2
\]
Thus, the solutions are \( X = 7 \) and \( X = 2 \).

### 2. \( 5X^2 - 6X + 1 = 0 \)
Again, we use the quadratic formula with \( a = 5 \), \( b = -6 \), and \( c = 1 \):
\[
X = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(5)(1)}}{2(5)} = \frac{6 \pm \sqrt{36 - 20}}{10} = \frac{6 \pm \sqrt{16}}{10}
\]
\[
X = \frac{6 \pm 4}{10}
\]
Thus, the two solutions are:
\[
X = \frac{6 + 4}{10} = 1 \quad \text{or} \quad X = \frac{6 - 4}{10} = 0.2
\]
The solutions are \( X = 1 \) and \( X = 0.2 \).

### 3. \( 6X^2 - 24 = 0 \)
This equation can be simplified by dividing the whole equation by 6:
\[
X^2 - 4 = 0
\]
This can be factored as:
\[
(X - 2)(X + 2) = 0
\]
Thus, the solutions are:
\[
X = 2 \quad \text{or} \quad X = -2
\]

### 4. \( 5X^2 + 10 = 0 \)
First, subtract 10 from both sides:
\[
5X^2 = -10
\]
Now, divide by 5:
\[
X^2 = -2
\]
Since the square of a real number cannot be negative, this equation has **no real solutions**. The solutions are complex:
\[
X = \pm \sqrt{-2} = \pm i\sqrt{2}
\]

### 5. \( 6X^2 - 9X = 0 \)
Factor out \( X \):
\[
X(6X - 9) = 0
\]
Set each factor to 0:
\[
X = 0 \quad \text{or} \quad 6X - 9 = 0
\]
Solve for \( X \) in the second equation:
\[
6X = 9 \quad \Rightarrow \quad X = \frac{9}{6} = \frac{3}{2}
\]
Thus, the solutions are \( X = 0 \) and \( X = \frac{3}{2} \).

---

If you need further details or have any questions, feel free to ask! Here are some related questions:

1. How do you determine the nature of solutions for quadratic equations?
2. Can you explain why some quadratic equations have complex solutions?
3. What is the discriminant, and how does it help in solving quadratic equations?
4. How would you solve quadratic equations by completing the square?
5. Can you graph these quadratic equations to visualize their solutions?

**Tip:** When using the quadratic formula, always check the discriminant \( b^2 - 4ac \) to know whether the solutions are real, complex, or repeated.

X^2 - 9X + 14 = 0, 5X^2 - 6X + 1 = 0, 6X^2 - 24 = 0, 5X^2 + 10 = 0, 6X^2 - 9X = 0

Explore how to solve various quadratic equations such as X^2 - 9X + 14 = 0 and 5X^2 - 6X + 1 = 0, using the quadratic formula and factoring techniques. Learn how to handle equations with real and complex solutions. Suitable for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation