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]
