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 set of values of a for which (a - 1)x^2 - (a + 1)x + a - 1 ≥ 0 is true for all x≥2

Solution

(c) Given,

(a - 1)x^2 - (a + 1)x + a - 1 ≥ 0

⇒ a(x^2 - x + 1) - (x^2 + x + 1) ≥ 0

⇒ a ≥ ( x^2 + x + 1 )/( x^2 - x + 1 )

= 1 + 2x/( x^2 - x + 1 )

= 1 + 2/( x + 1/x - 1) (1)

Let y = x + 1/x. Now, y is increasing in [2,∞). Hence,

1 + 2/( x + 1/x - 1)∈(1 , 7/3]

For all x ≥ 2, Eq. (1) should be true. Hence, a > 7/3

(c) Given,

(a - 1)x^2 - (a + 1)x + a - 1 ≥ 0

⇒ a(x^2 - x + 1) - (x^2 + x + 1) ≥ 0

⇒ a ≥ ( x^2 + x + 1 )/( x^2 - x + 1 )

= 1 + 2x/( x^2 - x + 1 )

= 1 + 2/( x + 1/x - 1) (1)

Let y = x + 1/x. Now, y is increasing in [2,∞). Hence,

1 + 2/( x + 1/x - 1)∈(1 , 7/3]

For all x ≥ 2, Eq. (1) should be true. Hence, a > 7/3

**Q2.**If w = Î± + iÎ², where Î² ≠ 0 and z ≠ 1, satisfies the condition that (( w - w ̅z )/( 1 - z ) ) is purely real, then the set of values of z is

**Q3.**The number of points of intersection of two curves y = 2 sinx and y = 5x^2 + 2x + 3 is

Solution

(a)

Minimum value of 5x^2 + 2x + 3 is

-D/4a = -( (2)^2 - 4(5)(3) )/( 4(5) ) >2

Where maximum value of 2 sinx is 2. Therefore, the two curves do not meet at all

(a)

Minimum value of 5x^2 + 2x + 3 is

-D/4a = -( (2)^2 - 4(5)(3) )/( 4(5) ) >2

Where maximum value of 2 sinx is 2. Therefore, the two curves do not meet at all

**Q5.**If one root of the equation ax^2 + bx + c = 0 is square of the other, then a( c - b )^3 = cX, where X is

Solution

(c)

If one root is square of the other root of the equation ax^2 + bx + c = 0, then

Î² = Î±^2 ⇒ Î±^2 + Î± = -b/a and Î±^2 a = c/a

By eliminating Î±, we get

b^3 + ac^2 + a^2 c = 3abc

Which can be written in the form a(c - b)^3 = c(a - b)^3

Alternative solution:

Let the roots be 2 and 4. Then the equation is x^2 - 6x + 8 = 0

Here obviously,

X = ( a(c - b)^3 )/c = ( 1(14)^3 )/8 = 14/2 × 14/2 × 14/2 = 7^3

Which is given by ( a - b )^3 = 7^3

(c)

If one root is square of the other root of the equation ax^2 + bx + c = 0, then

Î² = Î±^2 ⇒ Î±^2 + Î± = -b/a and Î±^2 a = c/a

By eliminating Î±, we get

b^3 + ac^2 + a^2 c = 3abc

Which can be written in the form a(c - b)^3 = c(a - b)^3

Alternative solution:

Let the roots be 2 and 4. Then the equation is x^2 - 6x + 8 = 0

Here obviously,

X = ( a(c - b)^3 )/c = ( 1(14)^3 )/8 = 14/2 × 14/2 × 14/2 = 7^3

Which is given by ( a - b )^3 = 7^3

**Q6.**Let x,y,z,t be real numbers x^2 + y^2 = 9 , z^2 + t^2 = 4 and xt - yz = 6. Then the greatest value of P = xz is

Solution

(b)

x = 3 cosÎ¸ ; y = 3sinÎ¸

z = 2cosÏ• ; t = 2sinÏ•

∴ 6cosÎ¸ sinÏ• - 6sinÎ¸ cos[Ï•=6]

⇒sin[(Ï• - Î¸)] = 1

⇒Ï• = 90° + Î¸

⇒P = xz = -6sinÎ¸ cosÎ¸ = -3sin2Î¸

⇒P_max = 3

(b)

x = 3 cosÎ¸ ; y = 3sinÎ¸

z = 2cosÏ• ; t = 2sinÏ•

∴ 6cosÎ¸ sinÏ• - 6sinÎ¸ cos[Ï•=6]

⇒sin[(Ï• - Î¸)] = 1

⇒Ï• = 90° + Î¸

⇒P = xz = -6sinÎ¸ cosÎ¸ = -3sin2Î¸

⇒P_max = 3

**Q7.**Let Î» ∈ R , the origin and the non-real roots of 2z^2 + 2z + Î» = 0 form the three vertices of an equilateral triangle in the Argand plane then Î» is

Solution

(b)

2z^2 + 2z + Î» = 0

Let the roots be z_1 , z_2. Then,

z_1 + z_2 = -1 and z_1 z_2 = Î»/2

0 , z_1 , z_2 form, an equilateral triangle

∴ z_1^2 + z_2^2 = z_1 z_2

⇒ (z_1 + z_2 )^2 = 3z_1 z_2

⇒ 1 = 3Î»/2

⇒ Î» = 2/3

(b)

2z^2 + 2z + Î» = 0

Let the roots be z_1 , z_2. Then,

z_1 + z_2 = -1 and z_1 z_2 = Î»/2

0 , z_1 , z_2 form, an equilateral triangle

∴ z_1^2 + z_2^2 = z_1 z_2

⇒ (z_1 + z_2 )^2 = 3z_1 z_2

⇒ 1 = 3Î»/2

⇒ Î» = 2/3

**Q8.**The number of values of k for which [x^2 - (k - 2)x +k^2 ]× [x^2 + kx + (2k - 1)] is a perfect square is

Solution

(b)

Verify by selecting particular values of a and b

Let a = -9 and b = 4. Then,

√a √b = √( -9 ) √4 =( 3i )( 2 ) = 6i

From option (a), we have

-√( |a|b ) = -√(|-9|×4) = -√36 = -6

From option (b), we have

√( |a|bi ) = √( |-9|×4 ) i = 6i

(b)

Verify by selecting particular values of a and b

Let a = -9 and b = 4. Then,

√a √b = √( -9 ) √4 =( 3i )( 2 ) = 6i

From option (a), we have

-√( |a|b ) = -√(|-9|×4) = -√36 = -6

From option (b), we have

√( |a|bi ) = √( |-9|×4 ) i = 6i

**Q9.**Let p(x) = 0 be a polynomial equation of the least possible degree, with rational coefficients, having ∛7+∛49 as one of its roots. Then the product of all the roots of p(x) = 0 is

Solution

(a)

x = ∛7 + ∛49

⇒ x^3 = 7 + 49 + 3∛7 ∛49 ( ∛7 + ∛49 ) = 56 + 21x

⇒ x^3 - 21x - 56 = 0

Therefore, the product of roots is 56

(a)

x = ∛7 + ∛49

⇒ x^3 = 7 + 49 + 3∛7 ∛49 ( ∛7 + ∛49 ) = 56 + 21x

⇒ x^3 - 21x - 56 = 0

Therefore, the product of roots is 56

**Q10.**The number of real solutions of the equation |x|^2 - 3|x| + 2 = 0 is

Solution

(a)

|x|^2 - 3|x| + 2 = 0

⇒ ( |x| - 2) (|x| - 1) = 0

⇒ |x| = 1 or 2

⇒x = ±1 , ±2

Hence, there are four real solutions

(a)

|x|^2 - 3|x| + 2 = 0

⇒ ( |x| - 2) (|x| - 1) = 0

⇒ |x| = 1 or 2

⇒x = ±1 , ±2

Hence, there are four real solutions