https://file.aiquickdraw.com/mathbot/file/1727966959704euzpvrhw.pdf

Let's break down and solve each of the questions from the quiz:

### 1. Solve the following equations by factorization:

#### a. \( x^2 - x = 12 \)
We rearrange the equation as:
\[
x^2 - x - 12 = 0
\]
Factorizing this quadratic equation:
\[
(x - 4)(x + 3) = 0
\]
Thus, the solutions are:
\[
x = 4 \quad \text{or} \quad x = -3
\]

#### b. \( x^2 + 30 = x \)
Rearrange the equation:
\[
x^2 - x + 30 = 0
\]
The discriminant \(\Delta = (-1)^2 - 4(1)(30) = 1 - 120 = -119\), so there are no real solutions, only complex ones.

#### c. \( 2x^2 + 6x = -16 \)
Rearrange the equation:
\[
2x^2 + 6x + 16 = 0
\]
Dividing by 2:
\[
x^2 + 3x + 8 = 0
\]
The discriminant is \(\Delta = 3^2 - 4(1)(8) = 9 - 32 = -23\), so there are no real solutions, only complex ones.

#### d. \( 4x^2 - 17 = x \)
Rearrange the equation:
\[
4x^2 - x - 17 = 0
\]
Using the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-17)}}{2(4)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 272}}{8} = \frac{1 \pm \sqrt{273}}{8}
\]
So, the solutions are:
\[
x = \frac{1 + \sqrt{273}}{8} \quad \text{or} \quad x = \frac{1 - \sqrt{273}}{8}
\]

### 2. Find the x-intercept (abscisse à l’origine) of the following equations:

#### a. \( 2x^2 + 3x - 40 = 0 \)
Using the quadratic formula:
\[
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-40)}}{2(2)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 320}}{4} = \frac{-3 \pm \sqrt{329}}{4}
\]
The solutions are:
\[
x = \frac{-3 + \sqrt{329}}{4} \quad \text{or} \quad x = \frac{-3 - \sqrt{329}}{4}
\]

#### b. \( 4x^2 + 8x - 12 = 0 \)
Dividing by 4:
\[
x^2 + 2x - 3 = 0
\]
Factorizing:
\[
(x - 1)(x + 3) = 0
\]
The solutions are:
\[
x = 1 \quad \text{or} \quad x = -3
\]

#### c. \( x^2 - 6x - 16 = 0 \)
Using the quadratic formula:
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-16)}}{2(1)}
\]
\[
x = \frac{6 \

Learn how to solve quadratic equations by factorization and using the quadratic formula. This guide covers examples like solving x^2 - x = 12, x^2 + 30 = x, and finding the x-intercepts of quadratic functions. Ideal for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation