## Complex Number & Quadratic Equation Quiz-7

Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..

Q1. The complex numbers z = x + iy which satisfy the equation |( z - 5i )/(z + 5i )| = 1 lie on
•  The x-axis
•  The straight line y = 5
•  A circle passing through the origin
•  None of these
Solution
(a)
We know that |z - z_1 | = |z - z_2 |. Then locus of z is the line, which is a perpendicular bisector of line segment joining z_1 and z_2
Hence,
z = x + iy
⇒ |z - 5i| = |z + 5i|
Therefore, z remains equidistant from z_1 = 5i and z_2 = 5i. Hence, z lies on perpendicular bisector of line segment joining z_1 and z_2, which is clearly the real axis or y = 0
Alternative solution:
|( z - 5i )/( z + 5i )| = 1
⇒ | x + iy - 5i| = |x + iy + 5i|
⇒ | x + (y - 5)i | = |x + (y + 5) i|
⇒ x^2 + (y - 5)^2 = x^2 + (y + 5)^2
⇒ x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25
⇒ 20y = 0
⇒ y = 0

Q2. If Î± and Î² (Î± < Î²) are the roots of the equation x^2 + bx + c = 0, where c<0
•  0 < Î± < Î²
•  Î± < 0 < Î² < |Î±|
•  Î± < Î² < 0
•  Î± < 0 < |Î±| < Î²
Solution
(b)
Here D = b^2 - 4c > 0 because c < 0 < b. So, roots are real and unequal. Now,
Î± + Î² = -b < 0 and Î±Î² = c < 0
Therefore, one root is positive and the other root is negative, the negative root being numerically bigger. As Î±<Î², so Î± is the negative root while Î² is the positive root. So, |Î±| > Î² and Î± < 0 < Î² < |Î±|

Q3.  All the values of m for which both the roots of the equation x^2 - 2mx + m^2 - 1 = 0 are greater than -2 but less than 4, lie in the interval
•  -2 < m < 0
•  m > 3
•  -1 < m < 3
•  1 < m < 4
Solution
(c)
The given equation is
x^2 - 2mx + m^2 - 1 = 0
⇒ (x - m)^2 - 1 = 0
⇒ (x - m + 1)(x - m - 1) = 0
⇒ x = m - 1 , m + 1
From given condition,
m - 1 > -2 and m + 1 < 4
⇒ m > -1 and m < 3
Hence, -1 < m < 3

Q4. Two towns A and B are 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school be built at
•  Town B
•  45 km from town A
•  Town A
•  45 km from town B
Solution
(c)
Let the distance of the school from A be x. Therefore, the distance of the school from B is 60 - x. The total distance covered by 200 students is
[150x + 50( 60 - x )] = [ 100x + 3000 ]
This is minimum when x = 0. Hence, the school should be at town A

Q5. If z = [ (√3/2) + i/2 ]^5 + [ (√3/2) - i/2 ]^5, then
•  Re(z) = 0
•  Im(z) = 0
•  Re(z) > 0 , Im(z) > 0
•  Re(z) > 0 , Im(z) < 0
Solution
(b)
z = (√3/2 + i/2)^5 + ( √3/2 - i/2)^5
= ( cos⁡[Ï€/6] + isin⁡[Ï€/6] )^5 + ( cos⁡[Ï€/6]- isin⁡[Ï€/6] )^5
= (cos⁡[5Ï€/6] + isin⁡[5Ï€/6]) + (cos⁡[5Ï€/6] - isin⁡[5Ï€/6] )
= 2cos⁡[5Ï€/6]
= -√3
⇒ Re(z) < 0 and Im(z) = 0
Alternative solution:
z = ¯z_1 + ¯z_2
Where
( √3/2 + i/2 )^5
⇒ z is real
⇒ Im(z) = 0

Q6. Let p and q be real numbers such that p ≠ 0 , p^3 ≠ q and p^3 ≠ -q. If Î± and Î² are non-zero complex numbers satisfying Î± + Î² = -p and Î±^3 + Î²^3 = q, then a quadratic equation having Î±/Î² and Î²/Î± as its roots is
•  (p^3 + q)x^2 - (p^3 + 2q)x + (p^3 + q) = 0
•  (p^3+q) x^2-(p^3-2q)x+(p^3 + q) = 0
• (p^3 - q)x^2 - (5p^3 - 2q)x + (p^3 - q) = 0
•  (p^3 - q)x^2 - ( [5p]^3 + 2q )x + (p^3 - q) = 0
Solution
(b)
Sum of roots = Î±/Î² + Î²/Î± = ( Î±^2 + Î²^2 )/Î±Î² and product = 1
Given, Î± + Î² = -p and Î±^3 + Î²^3 = q
⇒ (Î± + Î² )( Î±^2 - Î±Î² + Î²^2 ) = q
∴ Î±^2 + Î²^2 - Î±Î² = (-q)/p ...(i)
And (Î± + Î²)^2 = p^2
⇒ Î±^2 + Î²^2 + 2Î±Î² = p^2
From Eqs. (i) and (ii), we get
Î±^2 + Î²^2 = ( p^3 - 2q )/3p
And Î±Î² = ( p^3 + q )/3p
∴ Required equation is
x^2 - ( ( p^3 - 2q ) x)/( ( p^3 + q ) ) + 1 = 0
⇒ ( p^3 + q )x^2-( p^3 - 2q )x + ( p^3 + q ) = 0

Q7.  Let p and q be roots of the equation x^2 - 2x + A = 0 and let r and s be the roots of the equation x^2 - 18x + B = 0. If p < q < r < s are in arithmetic progression, then the values of A and B are
•  3 , -77
•  3 , 77
•  -3 , -77
•  -3 , 77
Solution

(d)
Let the four numbers in A.P. be p = a - 3d , q = a - d , r = a + d , s = a + 3d. Therefore,
p + q = 2 , r + s = 18
Given that pq = A , rs = B
∴ p + q + r + s = 4a = 20
⇒ a = 5
Now, p + q = 2 ⇒ 10 - 4d = 2
r + s = 18 ⇒ 10 + 4d = 18
∴ d = 2
Hence, the numbers are -1 , 3 , 7 , 11
pq = A = -3 , rs = B = 77
Q8. If Î± and Î² are the roots of the equation x^2 + px + q = 0, and Î±^4 and Î²^4 are the roots of x^2 - rx + q = 0, then the roots of x^2 - 4qx + 2q^2 - r = 0 are always
•  Both non-real
•  Both positive
•  Both negative
•  Opposite in sign
Solution
(d)
Î±,Î² are roots of x^2 + px + q = 0. Hence,
Î± + Î² = -p br Î±Î² = q
Now,
Î±^4 , Î²^4 are roots x^2 - px + q = 0. Hence,
Î±^4 + Î²^4 = r , Î±^4 Î²^4 = q
Now, for equation x^2 - 4qx + 2q^2 - r = 0, product of roots is
2q^2 - r = 2(Î±Î²)^2 - (Î±^4 + Î²^4)
= - (Î±^2 - Î²^2 )^2
< 0
As product of roots is negative, so the roots must be real

P ≡ (-1 , 0) , Q ≡ (-1 + √2 , √2)
R≡(-1 + √2 , -√2) , S ≡ (1 , 0) is represented by
•  |z + 1| > 2 , | arg⁡[(z+1)<Ï€/4] |
•  |z + 1| < 2 , arg⁡[(z + 1) < Ï€/2]
•  |z - 1| > 2 , arg⁡[(z + 1) > Ï€/4]
•  |z - 1| < 2 , |arg⁡[(z + 1) > Ï€/4]
Solution
(a)
Here, |PQ| = |PS| = |PR| = 2
∴ Shaded part represents the external part of circle having centre ( -1 , 0 ) and radius 2
As we know equation of circle having centre z_0 and radius r, is
|z - z_0 | = r
∴ |z — ( -1 + 0i )| > 2
⇒ |z + 1| > 2 …(i)
Also, argument of z + 1 with represent to positive direction of x-axis is Ï€/4
∴ arg⁡[(z + 1) ≤ Ï€/4 ]
And argument of z + 1 in a] = anti-clockwise direction is –Ï€/4.
∴ -Ï€/4 ≤ arg⁡(z + 1)
⇒|arg⁡(z + 1) | ≤ Ï€/4

Q10. Number of values of a for which equations x^3 + ax + 1 = 0 and x^4 + ax^2 + 1 = 0 have a common root
•  0
•  1
•  2
• Infinte
Solution
(b)
Given equations are
x^3 + ax + 1 = 0
or x^4 + ax^2 + x = 0 (1)
and x^4 + ax^2 + 1 = 0 (2)
From (1) – (2), we get x = 1. Thus, x = 1 is the common roots.
Hence,
1 + a + 1 = 0
⇒ a = -2

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