The Solution:
Given the equation below:
[tex]x^2-(k^2-3k)x+24=0[/tex]We are required to find the value of k that will make the sum of the solutions to be 10.
Step 1:
Let:
[tex]\begin{gathered} k^2-3k\text{ be represented with b} \\ \text{ So that we have} \\ k^2-3k=b\ldots eqn(1) \end{gathered}[/tex]So, the given equation becomes:
[tex]x^2-bx+24=0[/tex]We shall the Quadratic Formula Method to solve for x in terms of b.
In this case,
[tex]\begin{gathered} a=1 \\ b=-b \\ c=24 \end{gathered}[/tex]Substituting, we get
[tex]x=\frac{-b\pm\text{ }\sqrt[]{(-b)^2-(4\times1\times24)}}{2(1)}[/tex][tex]x=\frac{-b\pm\text{ }\sqrt[]{b^2-96}}{2}[/tex]So, the solutions to the given equation are:
[tex]\begin{gathered} x=\frac{-b+\text{ }\sqrt[]{b^2-96}}{2} \\ \text{ or} \\ x=\frac{-b-\text{ }\sqrt[]{b^2-96}}{2} \end{gathered}[/tex]Equating their sum to 10.
[tex]\begin{gathered} \frac{-b+\text{ }\sqrt[]{b^2-96}}{2}+\frac{-b-\text{ }\sqrt[]{b^2-96}}{2}=10 \\ \\ \\ \frac{-b+\text{ }\sqrt[]{b^2-96}+-b-\text{ }\sqrt[]{b^2-96}}{2}=10 \end{gathered}[/tex]Simplifying, we get
[tex]\begin{gathered} \frac{-2b}{2}=10 \\ \\ -b=10 \end{gathered}[/tex]Substituting for b, we get
[tex]\begin{gathered} -(k^2-3k)=10 \\ k^2-3k=-10 \\ k^2-3k+10=0 \end{gathered}[/tex]Solving for k by the Quadratic Formula method of solving quadratic equation, we get
[tex]k=\frac{-b\pm\text{ }\sqrt[]{b^2-4ac}}{2a}[/tex]Where
[tex]a=1,b=-3\text{ and c=10}[/tex]Substituting, we get
[tex]k=\frac{-(-3)\pm\text{ }\sqrt[]{(-3)^2-(4\times1\times10)}}{2(1)}[/tex][tex]k=\frac{3\pm\text{ }\sqrt[]{9^{}-40}}{2}=\frac{3\pm\text{ }\sqrt[]{-31}}{2}[/tex][tex]\begin{gathered} k=\frac{3+\text{ }\sqrt[]{-31}}{2}\text{ or }k=\frac{3-\text{ }\sqrt[]{-31}}{2} \\ \end{gathered}[/tex]Therefore, the correct answer is
[tex]k=\frac{3+\text{ }\sqrt[]{-31}}{2}\text{ or }k=\frac{3-\text{ }\sqrt[]{-31}}{2}[/tex]Alternatively,
We can use the sum of roots formula below:
[tex]\begin{gathered} \text{ Sum of roots = }\frac{-b}{a} \\ \text{if given a quadratic equation of the form ax}^2+bx+c=0 \end{gathered}[/tex]So, we get
[tex]\begin{gathered} a=1 \\ b=-(k^2-3k) \\ c=24 \end{gathered}[/tex]So,
[tex]\begin{gathered} \text{ Sum=}\frac{--(k^2-3k)}{1}=10 \\ \\ k^2-3k=10 \\ \\ k^2-3k-10=0 \end{gathered}[/tex]Then you can now solve from here as have done in the previous method.
Solve the quadratic equation above for k.
