Question

j133b obtain quadratic equation using 2 roots.
(6) \\(\frac{1+\sqrt{5}}{2}\\), \\(\frac{1-\sqrt{5}}{2}\\)
(7) \\(a\\), \\(b\\)
(8) \\(a+bi\\), \\(a-bi\\)
(9) \\(3a+2\sqrt{5}b\\), \\(3a-2\sqrt{5}b\\)

note: a quadratic equation with solutions \\(\frac{1+\sqrt{5}}{2}\\) and \\(\frac{1-\sqrt{5}}{2}\\) can be written
\\(\left(x-\frac{1+\sqrt{5}}{2}\
ight)\left(x-\frac{1-\sqrt{5}}{2}\
ight)=0\\).
this can be expanded and simplified; however it is simpler to find the quadratic equation using the method shown.

Explanation:

Step1: Identify the general formula

$$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$$

Step2: Solve question (6)

$$\text{Sum} = \frac{1+\sqrt{5}}{2} + \frac{1-\sqrt{5}}{2} = 1$$

$$\text{Product} = \frac{1+\sqrt{5}}{2} \cdot \frac{1-\sqrt{5}}{2} = \frac{1-5}{4} = -1$$

$$x^2 - x - 1 = 0$$

Step3: Solve question (7)

$$\text{Sum} = a + b$$

$$\text{Product} = ab$$

$$x^2 - (a + b)x + ab = 0$$

Step4: Solve question (8)

$$\text{Sum} = (a + bi) + (a - bi) = 2a$$

$$\text{Product} = (a + bi)(a - bi) = a^2 + b^2$$

$$x^2 - 2ax + (a^2 + b^2) = 0$$

Step5: Solve question (9)

$$\text{Sum} = (3a + 2\sqrt{5}b) + (3a - 2\sqrt{5}b) = 6a$$

$$\text{Product} = (3a + 2\sqrt{5}b)(3a - 2\sqrt{5}b) = 9a^2 - 20b^2$$

$$x^2 - 6ax + (9a^2 - 20b^2) = 0$$

Answer:

(6) \(x^2 - x - 1 = 0\)
(7) \(x^2 - (a + b)x + ab = 0\)
(8) \(x^2 - 2ax + (a^2 + b^2) = 0\)
(9) \(x^2 - 6ax + (9a^2 - 20b^2) = 0\)