Test: Quadratic Equations

x² - ix + 6 = 0
x² - 3ix + 2ix - 6i² = 0    { i² = -1}
x(x-3i) + 2i(x-3i) = 0
(x+2i) (x-3i) = 0
x = -2i, 3i 

{(1 + i)/(1 - i)}ⁿ = 1
multiply (1 + i) numerator as well as denominator .
{(1 + i)(1 + i)/(1 - i)(1 + i)}ⁿ = 1
{(1 + i)²/(1² - (i)²)}ⁿ = 1
{(1 + i² +2i)/2 }ⁿ = 1
{(2i)/2}ⁿ = 1
{i}ⁿ = 1
we know, i⁴ⁿ = 1 where , n is an integer.
so, n = 4n where n is an integers
e.g n = 4 { because least positive integer 1 }
hence, n = 4

x²−3x−2i=0,

or  x² + 3ix − 2=0      (divided by i)
 x= −3i ± [−9 + 4.1.2]^1/2/2.1
= −3i ±  (−1)^1/2/2
​= (−3i ± i)/2
=> x=−i or x=−2i

x² + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
⇒ x = 3 or x = -5.

2x² + 3ix + 2 = 0
Using quadratic equation;
we know, x = (-b ± √b² - 4ac)/2a
x =  [-3i ± √(3i)² - 4x2x2]/2x2
= -3i ± √-25/4
= i(-3±5)/4
x = i/2, -2i

2x² - 4x + 3 = 0
x = [-(-4) +- (√16-24)]/2(2)
x = (4 +- i√8)/4
x = (4 +- 2i√2)/(2 * √2 * √2)
x = 2(2 +- i√2)/(2 * √2 * √2)
x = 1 +- i/√2

Also, αβ = r/p, which is also rational. α + β = (a+√b) + (a-√b) = 2a, a rational number and, αβ = (a+√b)(a-√b) = a² - b, a rational number. So, the other root of a quadratic equation having the one root as (a+√b) is (a-√b), where a and b are rational numbers.

Let these three terms are, a−d,aanda+d.
Then,
a−d+a+a+d=51⇒3a=51⇒a=17
Now, (a−d)(a+d)=273
⇒(17−d)(17+d)=273
⇒289−d²=273⇒d²=16
⇒d=4 (As d can not be negative.)
So, third term will be, 17+4=21.

9x² + 16 = 0
9x² = -16
x² = -16/9
x = ± 4/3 i