Complex Numbers and Quadratic Equations
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New answer posted
a month agoContributor-Level 10
For the quadratic equation (k+1)tan²x - √2λ tanx + (k-1) = 0, the roots are tanα and tanβ.
Sum of roots: tanα + tanβ = √2λ / (k+1).
Product of roots: tanα tanβ = (k-1) / (k+1).
tan (α + β) = (tanα + tanβ) / (1 - tanα tanβ)
tan (α + β) = [√2λ / (k+1)] / [1 - (k-1)/ (k+1)]
tan (α + β) = [√2λ / (k+1)] / [ (k+1 - k + 1)/ (k+1)] = (√2λ) / 2 = λ/√2.
Given tan² (α+β) = 50.
(λ/√2)² = 50
λ²/2 = 50 ⇒ λ² = 100 ⇒ λ = ±10.
New answer posted
a month agoContributor-Level 10
Given Re (z-1)/ (2z+i) = 1, where z = x + iy.
(z-1)/ (2z+i) = [ (x-1) + iy] / [2x + I (2y+1)]
To rationalize, multiply the numerator and denominator by the conjugate of the denominator [2x - I (2y+1)].
Numerator = [ (x-1) + iy] * [2x - I (2y+1)] = 2x (x-1) - I (x-1) (2y+1) + i2xy + y (2y+1)
Real part of the numerator = 2x (x-1) + y (2y+1).
Denominator = (2x)² + (2y+1)².
Re (z-1)/ (2z+i) = [2x (x-1) + y (2y+1)] / [ (2x)² + (2y+1)²] = 1.
2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1.
0 = 2x² + 2y² + 2x + 3y + 1.
So, 2x² + 2y² + 2x + 3y + 1 = 0.
New answer posted
a month agoContributor-Level 9
log? /? [ (|z|+11)/ (|z|-1)²] < 2
(|z|+11)/ (|z|-1)² > (1/2)²
(|z|+11)/ (|z|-1)² > 1/4
⇒ 4|z| + 44 > |z|² - 2|z| + 1
⇒ |z|² - 6|z| - 43 < 0
⇒ |z| - 7 ≤ 0
∴ |z|max = 7
New answer posted
a month agoContributor-Level 10
The inequality is experience ( (|z|+3) (|z|-1) / (|z|+1) * log?2 ) ≥ log√? 16.
The right side is log? (1/2) 16 = log? (2? ¹) 2? = (4/-1)log?2 = -4. This seems incorrect.
Let's assume the base of the log on the right is √2. log√? 16 = log? (1/2) 2? = 2 * log?2? = 8.
The inequality becomes: 2^ (|z|+3) (|z|-1) / (|z|+1) ≥ 8 = 2³.
So, (|z|+3) (|z|-1) / (|z|+1) ≥ 3.
Let |z| = t. (t+3) (t-1) / (t+1) ≥ 3
t² + 2t - 3 ≥ 3t + 3
t² - t - 6 ≥ 0
(t-3) (t+2) ≥ 0
Since t = |z| ≥ 0, we must have t-3 ≥ 0.
So, t ≥ 3, which means |z| ≥ 3.
The minimum value of |z| is 3.
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
I = ∫[0 to 10] [sin(2πx)] / e^(x-[x]) dx.
The period of the integrand involves [sin(2πx)] which depends on the sign of sin(2πx) and {x} = x - [x], which has a period of 1.
Let f(x) = [sin(2πx)] / e^{x}.
The integral is ∫[0 to 10] f(x) dx = 10 * ∫[0 to 1] f(x) dx due to the periodicity of {x} and the integer period of sin(2πx).
In the interval (0, 1/2), sin(2πx) is between 0 and 1, so [sin(2πx)] = 0.
In the interval (1/2, 1), sin(2πx) is between -1 and 0, so [sin(2πx)] = -1.
At x=0, 1/2, 1, the value is 0.
So, ∫[0 to 1] f(x) dx = ∫[0 to 1/2] 0 dx + ∫[1/2 to 1] -1/e^x dx
= 0 + [-e^(-x) * (-1)] from 1/2 to 1 = [e^(-x)] from 1
New answer posted
a month agoContributor-Level 10
The general equation of a circle is given by:
az z? + α? z + αz? + d = 0
This can be rewritten as:
z? + (α? /a)z + (α/a)z? + d/a = 0
From this, we can identify the centre and radius:
Centre = -α/a
Radius = √ (|-α/a|² - d/a)
For a real circle to exist, the term under the square root must be non-negative:
|-α/a|² - d/a ≥ 0
|α|²/|a|² - d/a ≥ 0
|α|² - ad ≥ 0, where a ∈ R - {0}.
New answer posted
a month agoContributor-Level 10
∴ x² = |x|² = t let
9t² - 18t + 5 = 0
(3t - 1) (3t - 5) = 0
|x| = 1/3, 5/3
Product of roots = (1/3) (-1/3) (5/3) (-5/3) = 25/81
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