Question

j 138b

1. given that (x^2 - 12x + k = 0) has roots (alpha) and (2alpha), find the value of (alpha), then find the value of (k).

1. given that (x^2 + kx + 12 = 0) has roots (alpha) and (2alpha), find the value of (alpha), then find the value of (k).

Explanation:

Step 1: Identify relations for question 2

$$\alpha + 2\alpha = 12 \implies 3\alpha = 12$$

Step 2: Solve for \(\alpha\) in question 2

$$\alpha = 4$$

Step 3: Solve for \(k\) in question 2

$$k = \alpha \cdot 2\alpha = 2\alpha^2 = 2(4)^2 = 32$$

Step 4: Identify relations for question 3

$$\alpha \cdot 2\alpha = 12 \implies 2\alpha^2 = 12$$

Step 5: Solve for \(\alpha\) in question 3

$$\alpha^2 = 6 \implies \alpha = \pm\sqrt{6}$$

Step 6: Solve for \(k\) in question 3

$$-k = \alpha + 2\alpha = 3\alpha \implies k = -3\alpha = \mp 3\sqrt{6}$$

Answer:

For question 2: \(\alpha = 4\), \(k = 32\)
For question 3: \(\alpha = \pm \sqrt{6}\), \(k = \mp 3\sqrt{6}\) (or \(\alpha = \sqrt{6}, k = -3\sqrt{6}\) and \(\alpha = -\sqrt{6}, k = 3\sqrt{6}\))