Find the best solutions to your problems with the help of IDNLearn.com's experts. Get accurate and comprehensive answers to your questions from our community of knowledgeable professionals.
Sagot :
Sure, let's go through each part of the problem step-by-step.
### 1. Solutions of [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex]
To solve the equation [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex], we start by setting the equations equal to each other and solving for [tex]\( x \)[/tex].
[tex]\[ x^2 - 2x - 4 = -3x + 9 \][/tex]
We bring all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 4 + 3x - 9 = 0 \][/tex]
[tex]\[ x^2 + x - 13 = 0 \][/tex]
Solving this quadratic equation, we find the solutions:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Thus, the solutions for the intersection points are:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
### 2. [tex]\( y \)[/tex]-Coordinates of the [tex]\( y \)[/tex]-Intercepts
The [tex]\( y \)[/tex]-intercept is found by setting [tex]\( x = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ y = 0^2 - 2 \cdot 0 - 4 \][/tex]
[tex]\[ y = -4 \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ y = -3 \cdot 0 + 9 \][/tex]
[tex]\[ y = 9 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercepts are:
[tex]\[ y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \][/tex]
[tex]\[ y_{\text{intercept of } y = -3x + 9} = 9 \][/tex]
### 3. [tex]\( x \)[/tex]-Coordinates of the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercept is found by setting [tex]\( y = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
Solving the above quadratic equation, we get:
[tex]\[ x = 1 - \sqrt{5} \][/tex]
[tex]\[ x = 1 + \sqrt{5} \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ 0 = -3x + 9 \][/tex]
[tex]\[ 3x = 9 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, \; 1 + \sqrt{5} \][/tex]
[tex]\[ x_{\text{intercept of } y = -3x + 9} = 3 \][/tex]
### 4. [tex]\( y \)[/tex]-Coordinates of the Intersection Points
We have already found the [tex]\( x \)[/tex]-coordinates of the intersection points as:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
To find the [tex]\( y \)[/tex]-coordinates of these intersection points, we substitute these [tex]\( x \)[/tex]-values back into either of the original equations (since they are equal at the intersection points):
For [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex]:
[tex]\[ y_1 = \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 - 2 \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right) - 4 \][/tex]
For [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]:
[tex]\[ y_2 = \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 - 2 \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right) - 4 \][/tex]
So, the [tex]\( y \)[/tex]-coordinates of the intersection points are:
[tex]\[ y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \][/tex]
[tex]\[ y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \][/tex]
### 5. [tex]\( x \)[/tex]-Coordinates of the Intersection Points
The solutions are found in step 1:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Putting it all together, the results are as follows:
- [tex]\( x \)[/tex]-Coordinates of intersection points: [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex], [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]
- [tex]\( y \)[/tex]-Intercepts: [tex]\( y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \)[/tex], [tex]\( y_{\text{intercept of } y = -3x + 9} = 9 \)[/tex]
- [tex]\( x \)[/tex]-Intercepts: [tex]\( x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, 1 + \sqrt{5} \)[/tex], [tex]\( x_{\text{intercept of } y = -3x + 9} = 3 \)[/tex]
- [tex]\( y \)[/tex]-Coordinates of intersection points: [tex]\( y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \)[/tex], [tex]\( y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \)[/tex]
### 1. Solutions of [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex]
To solve the equation [tex]\( x^2 - 2x - 4 = -3x + 9 \)[/tex], we start by setting the equations equal to each other and solving for [tex]\( x \)[/tex].
[tex]\[ x^2 - 2x - 4 = -3x + 9 \][/tex]
We bring all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 4 + 3x - 9 = 0 \][/tex]
[tex]\[ x^2 + x - 13 = 0 \][/tex]
Solving this quadratic equation, we find the solutions:
[tex]\[ x = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Thus, the solutions for the intersection points are:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
### 2. [tex]\( y \)[/tex]-Coordinates of the [tex]\( y \)[/tex]-Intercepts
The [tex]\( y \)[/tex]-intercept is found by setting [tex]\( x = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ y = 0^2 - 2 \cdot 0 - 4 \][/tex]
[tex]\[ y = -4 \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ y = -3 \cdot 0 + 9 \][/tex]
[tex]\[ y = 9 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercepts are:
[tex]\[ y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \][/tex]
[tex]\[ y_{\text{intercept of } y = -3x + 9} = 9 \][/tex]
### 3. [tex]\( x \)[/tex]-Coordinates of the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercept is found by setting [tex]\( y = 0 \)[/tex] in the equation:
For [tex]\( y = x^2 - 2x - 4 \)[/tex]:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
Solving the above quadratic equation, we get:
[tex]\[ x = 1 - \sqrt{5} \][/tex]
[tex]\[ x = 1 + \sqrt{5} \][/tex]
For [tex]\( y = -3x + 9 \)[/tex]:
[tex]\[ 0 = -3x + 9 \][/tex]
[tex]\[ 3x = 9 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, \; 1 + \sqrt{5} \][/tex]
[tex]\[ x_{\text{intercept of } y = -3x + 9} = 3 \][/tex]
### 4. [tex]\( y \)[/tex]-Coordinates of the Intersection Points
We have already found the [tex]\( x \)[/tex]-coordinates of the intersection points as:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
To find the [tex]\( y \)[/tex]-coordinates of these intersection points, we substitute these [tex]\( x \)[/tex]-values back into either of the original equations (since they are equal at the intersection points):
For [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex]:
[tex]\[ y_1 = \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 - 2 \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right) - 4 \][/tex]
For [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]:
[tex]\[ y_2 = \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 - 2 \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right) - 4 \][/tex]
So, the [tex]\( y \)[/tex]-coordinates of the intersection points are:
[tex]\[ y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \][/tex]
[tex]\[ y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \][/tex]
### 5. [tex]\( x \)[/tex]-Coordinates of the Intersection Points
The solutions are found in step 1:
[tex]\[ x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \][/tex]
Putting it all together, the results are as follows:
- [tex]\( x \)[/tex]-Coordinates of intersection points: [tex]\( x_1 = -\frac{1}{2} + \frac{\sqrt{53}}{2} \)[/tex], [tex]\( x_2 = -\frac{\sqrt{53}}{2} - \frac{1}{2} \)[/tex]
- [tex]\( y \)[/tex]-Intercepts: [tex]\( y_{\text{intercept of } y = x^2 - 2x - 4} = -4 \)[/tex], [tex]\( y_{\text{intercept of } y = -3x + 9} = 9 \)[/tex]
- [tex]\( x \)[/tex]-Intercepts: [tex]\( x_{\text{intercepts of } y = x^2 - 2x - 4} = 1 - \sqrt{5}, 1 + \sqrt{5} \)[/tex], [tex]\( x_{\text{intercept of } y = -3x + 9} = 3 \)[/tex]
- [tex]\( y \)[/tex]-Coordinates of intersection points: [tex]\( y_1 = -\sqrt{53} - 3 + \left( -\frac{1}{2} + \frac{\sqrt{53}}{2} \right)^2 \)[/tex], [tex]\( y_2 = -3 + \sqrt{53} + \left( -\frac{\sqrt{53}}{2} - \frac{1}{2} \right)^2 \)[/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.
