solving quadratic equations using quadratic formula

1) 2X= - X+8
2) P²-9P+14=0

solve number 1 and 2, give me a formula

To solve quadratic equations using the quadratic formula, we use the general form of the quadratic equation:

\[
ax^2 + bx + c = 0
\]

The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Now let's solve the equations one by one.

### 1) \( 2X = -X + 8 \)

First, we need to rearrange the equation to match the general form \(ax^2 + bx + c = 0\).

\[
2X + X - 8 = 0
\]

Simplifying:

\[
3X - 8 = 0
\]

This equation is linear, not quadratic, so the solution can be found by isolating \(X\):

\[
X = \frac{8}{3}
\]

So, the solution for this equation is:

\[
X = \frac{8}{3}
\]

### 2) \( P^2 - 9P + 14 = 0 \)

This equation is already in the standard quadratic form where:
- \(a = 1\)
- \(b = -9\)
- \(c = 14\)

Using the quadratic formula:

\[
P = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)}
\]

Simplify the expression:

\[
P = \frac{9 \pm \sqrt{81 - 56}}{2}
\]

\[
P = \frac{9 \pm \sqrt{25}}{2}
\]

\[
P = \frac{9 \pm 5}{

Learn how to solve quadratic equations using the quadratic formula with detailed step-by-step solutions. Example problems include 2X = -X + 8 and P^2 - 9P + 14 = 0. Suitable for students in Grades 7-9.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation