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Complex Numbers and Quadratic Equations Ncert Solutions Maths class 11th Maths Class 11th

If |z + 1| = αz + β (i + 1) and $z = \frac{1}{2}$ – 2i, find α + β.

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alok kumar singh | Contributor-Level 10

3 weeks ago

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ .(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a = 1

b = 2

-> a + b = 3

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alok kumar singh

$z^{2} = - \bar{i z}$

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Given x² – 70x + λ = 0 with positive roots a and b where one of the root is less than 10 and $\frac{λ}{2}$ and $\frac{λ}{3}$ are not integers then find value of $\frac{\sqrt[]{α - 1} + \sqrt[]{β - 1}}{| α - β |}$ is equal to

alok kumar singh

Given : x² – 70x + l = 0

->Let roots be a and b

->b = 70 – a

->= a (70 – a)

l is not divisible by 2 and 3

->a = 5, b = 65

-> $\frac{\sqrt[]{5 - 1} + \sqrt[]{65 - 1}}{| 60 |} = | \frac{4 + 8}{60} | = \frac{1}{5}$

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alok kumar singh

z₁ + z₂ = 5

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$z_{1}^{3} + z_{2}^{3} = 125 - 3 z_{1} \cdot z_{2} (5)$

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⇒ 3z₁z₂ = 25 – 4 – 3i

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The integer ‘k’ for which the inequality x² – 2(3k – 1) x + 8k² – 7 > 0 is valid for every x in R, is:

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$x^{2} - 2 (3 k - 1) x + 8 k^{2} - 7 > 0 \Rightarrow a > 0, & D < 0,$

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Let z be those complex numbers which satisfy

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alok kumar singh

$G i v e n | z + 5 | \leq 4 . . . . . . . . . . (i)$

->Represent a circle ${(x + 5)}^{2} + y^{2} \leq 16$

$= z (1 + i) + \bar{z} (1 - i) \geq - 10 . . . . . . . . . . (i i)$

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