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x^{2}  ix + 6 = 0
x^{2}  3ix + 2ix  6i^{2} = 0 { i^{2} = 1}
x(x3i) + 2i(x3i) = 0
(x+2i) (x3i) = 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, i^{4n} = 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
x^{2}−3x−2i=0,
or x^{2 }+ 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
Find the roots of the quadratic equation: x^{2} + 2x  15 = 0?
x^{2} + 5x  3x  15 = 0
x(x + 5)  3(x + 5) = 0
(x  3)(x + 5) = 0
⇒ x = 3 or x = 5.
The solution of the quadratic equation: 2x^{2} + 3ix + 2 = 0
2x^{2} + 3ix + 2 = 0
Using quadratic equation;
we know, x = (b ± √b^{2}  4ac)/2a
x = [3i ± √(3i)^{2}  4x2x2]/2x2
= 3i ± √25/4
= i(3±5)/4
x = i/2, 2i
The solution of the quadratic equation : 2x^{2} – 4x + 3 = 0
2x^{2}  4x + 3 = 0
x = [(4) + (√1624)]/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
If one of the root of a quadratic equation with rational coefficients is rational, then other root must be
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.
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is
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−d^{2}=273⇒d^{2}=16
⇒d=4 (As d can not be negative.)
So, third term will be, 17+4=21.
9x^{2} + 16 = 0
9x^{2} = 16
x^{2} = 16/9
x = ± 4/3 i
156 videos176 docs132 tests

156 videos176 docs132 tests
