In the expansion of ( )( )6

2

1 2 1 1 k x k x x + + + , the coefficients of x and 2

x are 8 and 30 respectively. Find

the values of 1 k and 2

k.

Given the expansion of \((1 + k_1x)(1 + k_2x)^6\), we are tasked with finding the values of \(k_1\) and \(k_2\) such that the coefficients of \(x\) and \(x^2\) are 8 and 30, respectively.

### Step 1: Write the expansion for the first few terms

Expand \((1 + k_1x)(1 + k_2x)^6\) using the binomial theorem for \((1 + k_2x)^6\):

\[
(1 + k_2x)^6 = 1 + 6k_2x + 15k_2^2x^2 + \dots
\]

Now multiply this by \((1 + k_1x)\):

\[
(1 + k_1x)(1 + 6k_2x + 15k_2^2x^2 + \dots)
\]

### Step 2: Collect terms involving \(x\) and \(x^2\)

- The term involving \(x\) comes from:
  \[
  1 \cdot 6k_2x + k_1x \cdot 1 = (6k_2 + k_1)x
  \]
  The coefficient of \(x\) is \(6k_2 + k_1 = 8\).

- The term involving \(x^2\) comes from:
  \[
  1 \cdot 15k_2^2x^2 + k_1x \cdot 6k_2x = (15k_2^2 + 6k_1k_2)x^2
  \]
  The coefficient of \(x^2\) is \(15k_2^2 + 6k_1k_2 = 30\).

### Step 3: Solve the system of equations

We now have two equations:
1. \(6k_2 + k_1 = 8\)
2. \(15k_2^2 + 6k_1k_2 = 30\)

#### From equation (1):
Solve for \(k_1\):
\[
k_1 = 8 - 6k_2
\]

#### Substitute this into equation (2):

\[
15k_2^2 + 6(8 - 6k_2)k_2 = 30
\]

Simplify:
\[
15k_2^2 + 48k_2 - 36k_2^2 = 30
\]
\[
-21k_2^2 + 48k_2 - 30 = 0
\]

Divide through by -3 to simplify:
\[
7k_2^2 - 16k_2 + 10 = 0
\]

### Step 4: Solve the quadratic equation

Solve using the quadratic formula:
\[
k_2 = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(7)(10)}}{2(7)}
\]
\[
k_2 = \frac{16 \pm \sqrt{256 - 280}}{14}
\]
\[
k_2 = \frac{16 \pm \sqrt{-24}}{14}
\]

The discriminant is negative, indicating no real solution for \(k_2\). Please recheck the values given or the setup of the problem.

Would you like me to explain this in more detail or check for any possible corrections?

---

Here are 5 questions based on this problem:
1. How does the binomial theorem apply to expanding expressions like \((1 + k_2x)^6\)?
2. Can you describe how coefficients of specific terms like \(x\) or \(x^2\) are identified during expansion?
3. Why is the quadratic formula used to solve for \(k_2\), and what happens when there is a negative discriminant?
4. How would the process change if the discriminant were positive?
5. What are the implications of having no real solutions for this system of equations?

**Tip:** Always check the discriminant when solving quadratic equations to predict whether you'll get real, repeated, or complex roots.

In the expansion of ( )( )6 2 1 2 1 1 k x k x x + + + , the coefficients of x and 2 x are 8 and 30 respectively. Find the values of 1 k and 2 k.

Explore how to solve for k1 and k2 in the binomial expansion (1 + k1x)(1 + k2x)^6, where the coefficients of x and x^2 are given as 8 and 30, respectively. This detailed solution involves binomial expansion and solving a quadratic equation, ideal for students in Grades 10-12.

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