Question

For what values of k will the sum of the solutions of x^2 - (k^2 - 3k)x + 24=0 be 10?

Answer 1

The Solution:

Given the equation below:

`x^2-(k^2-3k)x+24=0`

We are required to find the value of k that will make the sum of the solutions to be 10.

Step 1:

Let:

`\begin{gathered} k^2-3k\text{ be represented with b} \\ \text{ So that we have} \\ k^2-3k=b\ldots eqn(1) \end{gathered}`

So, the given equation becomes:

`x^2-bx+24=0`

We shall the Quadratic Formula Method to solve for x in terms of b.

In this case,

`\begin{gathered} a=1 \\ b=-b \\ c=24 \end{gathered}`

Substituting, we get

`x=\frac{-b\pm\text{ }\sqrt[]{(-b)^2-(4*1*24)}}{2(1)}`
`x=\frac{-b\pm\text{ }\sqrt[]{b^2-96}}{2}`

So, the solutions to the given equation are:

`\begin{gathered} x=\frac{-b+\text{ }\sqrt[]{b^2-96}}{2} \\ \text{ or} \\ x=\frac{-b-\text{ }\sqrt[]{b^2-96}}{2} \end{gathered}`

Equating their sum to 10.

`\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}`

Simplifying, we get

`\begin{gathered} (-2b)/(2)=10 \\ \\ -b=10 \end{gathered}`

Substituting for b, we get

`\begin{gathered} -(k^2-3k)=10 \\ k^2-3k=-10 \\ k^2-3k+10=0 \end{gathered}`

Solving for k by the Quadratic Formula method of solving quadratic equation, we get

`k=\frac{-b\pm\text{ }\sqrt[]{b^2-4ac}}{2a}`

Where

`a=1,b=-3\text{ and c=10}`

Substituting, we get

`k=\frac{-(-3)\pm\text{ }\sqrt[]{(-3)^2-(4*1*10)}}{2(1)}`
`k=\frac{3\pm\text{ }\sqrt[]{9^{}-40}}{2}=\frac{3\pm\text{ }\sqrt[]{-31}}{2}`
`\begin{gathered} k=\frac{3+\text{ }\sqrt[]{-31}}{2}\text{ or }k=\frac{3-\text{ }\sqrt[]{-31}}{2} \\ \end{gathered}`

Therefore, the correct answer is

`k=\frac{3+\text{ }\sqrt[]{-31}}{2}\text{ or }k=\frac{3-\text{ }\sqrt[]{-31}}{2}`

Alternatively,

We can use the sum of roots formula below:

`\begin{gathered} \text{ Sum of roots = }(-b)/(a) \\ \text{if given a quadratic equation of the form ax}^2+bx+c=0 \end{gathered}`

So, we get

`\begin{gathered} a=1 \\ b=-(k^2-3k) \\ c=24 \end{gathered}`

So,

`\begin{gathered} \text{ Sum=}(--(k^2-3k))/(1)=10 \\ \\ k^2-3k=10 \\ \\ k^2-3k-10=0 \end{gathered}`

Then you can now solve from here as have done in the previous method.

Solve the quadratic equation above for k.