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Let α , β , be the roots of the equation x2−4λx+5=0  and  α,γ be the roots of the equation x2−(32+23)x+7+3λ3=0, If β+γ=32, then (α+2β+γ)2 is equal to...............

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    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago

    α , β are roots of x2 - 4 λ x + 5 = 0  

    ∴ α + β = 4 λ   a n d     α β = 5               

    Also α , γ  are roots of

    x 2 − ( 3 2 + 2 3 ) x + 7 + 3 3 λ = 0 , λ > 0

    ∴ α + γ = 3 2 + 2 3 , α γ = 7 + 3 2 λ               

    ? α is common root

    ∴ α 2 − 4 λ α + 5 = 0     …….(i)

    And

    α 2 − ( 3 2 + 2 3 ) α + 7 + 3 3 λ = 0      ….(ii)

    From (i) – (ii) ; we get

    α = 2 + 3 3 λ 3 2 + 2 3 − 4 λ               

    ? β + γ = 3 2               

    ∴ ( α + 2 β + γ ) 2 = ( α + β + β + γ ) 2               

    = 9 8             

Similar Questions for you

Let the lines (2 – i) z = (2 + i) z ¯     a n d     ( 2 + i ) z + ( i − 2 ) z ¯ − 4 i = 0 , (here i2 = -1) be normal to a circle C. If the line iz + z ¯ + 1 + i = 0  is tangent to this circle C, then its radius is:
V
Vishal Baghel

(2 – i) z = (2 + i) z ¯  , put z = x + iy

y = x 2 . . . . . . . . ( i )

(ii) ( 2 + i ) z + ( i − 2 ) z ¯ − 4 i = 0

x + 2y = 2

(iii) i z + z ¯ + 1 + i = 0

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

x = 1 . y = 1 2 ⇒ c e n t r e ( 1 , 1 2 )

Perpendicular distance of point ( 1 , 1 2 )  from x – y + 1 = 0 is p = r ⇒ | 1 − 1 2 + 1 2 | = r

r = 3 2 2

A quadratic function f(x) satisfies f(x) ≥ 0 for all real x. If f(2) = 0 and f(4) = 12, then the value of f(6) is :
V
Vishal Baghel

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 × 4² = 48

The area of the triangle vertices A(z), B(iz) and C(z + iz) is:
V
Vishal Baghel

Kindly consider the following figure

Let z be those complex numbers which satisfy

| z + 5 | ≤ 4     a n d     z ( 1 + i ) + z ¯ ( 1 − i ) ≥ − 1 0 ,     i = − 1 .

If the maximum value of | z + 1 | 2 i s     α + β 2 , then the value of (α + β) is…….
V
Vishal Baghel

G i v e n     | z + 5 | ≤ 4 . . . . . . . . . . ( i )

->Represent a circle ( x + 5 ) 2 + y 2 ≤ 1 6

= z ( 1 + i ) + z ¯ ( 1 − i ) ≥ − 1 0 . . . . . . . . . . ( i i )

->Represent a line X – y ≥ − 5

So max |z + 1|2 = AQ2

= ( − 4 − 2 2 ) 2 + 8

= 3 2 + 1 6 2 = α + β 2  

Hence α + β) = 48

Let z = 1 − i 3 2 , i = − 1 .  Then the value of

2 1 + ( z + 1 z ) 3 + ( z 2 + 1 z 2 ) 3 + ( z 3 + 1 z 3 ) 3 + . . . . . + ( z 2 1 + 1 z 2 1 ) 3 i s _ _ _ _ _ _ _ _ _ .

V
Vishal Baghel

S = 2 1 + ∑ k = 1 2 1 ( z k + 1 z k ) 3 = ∑ ( z k + z ¯ k ) 3 = 8 ∑ ( R e ( z k ) ) 3

( A s | z | = 1 ⇒ 1 z = z ¯ )

S = 2 ∑ k = 1 2 1 c o s k π + 6 ∑ k = 1 2 1 c o s k π 3 + 2 1

= 2 ( − 1 ) + 6 R e ( α + α 2 + α 3 ) + 2 1

Where α = e i ( 2 π / 6 )

= 2 ( − 1 ) + 6 ( 1 2 − 1 2 − 1 ) + 2 1

= − 8 + 2 1 = 1 3

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