Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Certainly! Let's solve the given system of equations:
1. [tex]\( 3655 = -k^2 + 168k - 152 \)[/tex]
2. [tex]\( k^2 = 168k + 3503 \)[/tex]
Step 1: Simplify the equations
First, let's rearrange each equation to get them into a standard polynomial form:
For the first equation:
[tex]\[ 3655 = -k^2 + 168k - 152 \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ 0 = -k^2 + 168k - 152 - 3655 \][/tex]
Combine the constants:
[tex]\[ 0 = -k^2 + 168k - 3807 \][/tex]
So, our first equation in standard form is:
[tex]\[ -k^2 + 168k - 3807 = 0 \][/tex]
For the second equation:
[tex]\[ k^2 = 168k + 3503 \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ k^2 - 168k - 3503 = 0 \][/tex]
Now we have two quadratic equations:
1. [tex]\[ -k^2 + 168k - 3807 = 0 \][/tex]
2. [tex]\[ k^2 - 168k - 3503 = 0 \][/tex]
Step 2: Solve the quadratic equations
Let's solve the first quadratic equation:
[tex]\[ -k^2 + 168k - 3807 = 0 \][/tex]
This quadratic equation can be rewritten as:
[tex]\[ k^2 - 168k + 3807 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -168 \)[/tex], and [tex]\( c = 3807 \)[/tex]. Plugging these into the quadratic formula gives:
[tex]\[ k = \frac{-(-168) \pm \sqrt{(-168)^2 - 4 \cdot 1 \cdot 3807}}{2 \cdot 1} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{28224 - 15228}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{12996}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm 114}{2} \][/tex]
This gives us the solutions:
[tex]\[ k = \frac{168 + 114}{2} = 141 \][/tex]
[tex]\[ k = \frac{168 - 114}{2} = 27 \][/tex]
So, for the first equation, the solutions are:
[tex]\[ k = 27 \text{ or } k = 141 \][/tex]
Next, let's solve the second quadratic equation:
[tex]\[ k^2 - 168k - 3503 = 0 \][/tex]
Again, we use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = -168 \)[/tex], and [tex]\( c = -3503 \)[/tex]:
[tex]\[ k = \frac{-(-168) \pm \sqrt{(-168)^2 - 4 \cdot 1 \cdot (-3503)}}{2 \cdot 1} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{28224 + 14012}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{42236}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{10559} \cdot \sqrt{4}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm 2\sqrt{10559}}{2} \][/tex]
[tex]\[ k = 84 \pm \sqrt{10559} \][/tex]
So, for the second equation, the solutions are:
[tex]\[ k = 84 - \sqrt{10559} \text{ or } k = 84 + \sqrt{10559} \][/tex]
Conclusion:
The solution to the system of equations [tex]\( 3655 = -k^2 + 168k - 152 \)[/tex] and [tex]\( k^2 = 168k + 3503 \)[/tex] are:
For the first equation:
[tex]\[ k = 27 \text{ or } k = 141 \][/tex]
For the second equation:
[tex]\[ k = 84 - \sqrt{10559} \text{ or } k = 84 + \sqrt{10559} \][/tex]
1. [tex]\( 3655 = -k^2 + 168k - 152 \)[/tex]
2. [tex]\( k^2 = 168k + 3503 \)[/tex]
Step 1: Simplify the equations
First, let's rearrange each equation to get them into a standard polynomial form:
For the first equation:
[tex]\[ 3655 = -k^2 + 168k - 152 \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ 0 = -k^2 + 168k - 152 - 3655 \][/tex]
Combine the constants:
[tex]\[ 0 = -k^2 + 168k - 3807 \][/tex]
So, our first equation in standard form is:
[tex]\[ -k^2 + 168k - 3807 = 0 \][/tex]
For the second equation:
[tex]\[ k^2 = 168k + 3503 \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ k^2 - 168k - 3503 = 0 \][/tex]
Now we have two quadratic equations:
1. [tex]\[ -k^2 + 168k - 3807 = 0 \][/tex]
2. [tex]\[ k^2 - 168k - 3503 = 0 \][/tex]
Step 2: Solve the quadratic equations
Let's solve the first quadratic equation:
[tex]\[ -k^2 + 168k - 3807 = 0 \][/tex]
This quadratic equation can be rewritten as:
[tex]\[ k^2 - 168k + 3807 = 0 \][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -168 \)[/tex], and [tex]\( c = 3807 \)[/tex]. Plugging these into the quadratic formula gives:
[tex]\[ k = \frac{-(-168) \pm \sqrt{(-168)^2 - 4 \cdot 1 \cdot 3807}}{2 \cdot 1} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{28224 - 15228}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{12996}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm 114}{2} \][/tex]
This gives us the solutions:
[tex]\[ k = \frac{168 + 114}{2} = 141 \][/tex]
[tex]\[ k = \frac{168 - 114}{2} = 27 \][/tex]
So, for the first equation, the solutions are:
[tex]\[ k = 27 \text{ or } k = 141 \][/tex]
Next, let's solve the second quadratic equation:
[tex]\[ k^2 - 168k - 3503 = 0 \][/tex]
Again, we use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = -168 \)[/tex], and [tex]\( c = -3503 \)[/tex]:
[tex]\[ k = \frac{-(-168) \pm \sqrt{(-168)^2 - 4 \cdot 1 \cdot (-3503)}}{2 \cdot 1} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{28224 + 14012}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{42236}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm \sqrt{10559} \cdot \sqrt{4}}{2} \][/tex]
[tex]\[ k = \frac{168 \pm 2\sqrt{10559}}{2} \][/tex]
[tex]\[ k = 84 \pm \sqrt{10559} \][/tex]
So, for the second equation, the solutions are:
[tex]\[ k = 84 - \sqrt{10559} \text{ or } k = 84 + \sqrt{10559} \][/tex]
Conclusion:
The solution to the system of equations [tex]\( 3655 = -k^2 + 168k - 152 \)[/tex] and [tex]\( k^2 = 168k + 3503 \)[/tex] are:
For the first equation:
[tex]\[ k = 27 \text{ or } k = 141 \][/tex]
For the second equation:
[tex]\[ k = 84 - \sqrt{10559} \text{ or } k = 84 + \sqrt{10559} \][/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
