A mathematical equation having a complex number comprises of the real and imaginary sections. Complex numbers are nothing but a combination of two numbers (real, imaginary)..

**Q1.**If the roots of the equation ax^2 + bx +c =0 are of the form (k+1)/k and (k+2)/(k+1) , then (a+b+c)^2 is equal to

Solution

(c)

We have,

(k + 1)/k + (k + 2)/(k + 1) = -b/a (1)

and (k + 1)/k + (k + 2)/(k + 1) = c/a

⇒ (k + 2)/k = c/a or 2/k = c/a - 1 = (c - a)/a or k = 2a/(c - a) (2)

Now, eliminate k. Putting the value of k in Eq. (1), we get

(c + a)/2a + 2c/(c + a) = -b/a

⇒(c + a )^2 + 4ac = - 2b( a + c)

⇒( a + c )^2 + 2b( a + c) = -4ac

Adding b^2 to both sides, we have

(a + b + c )^2 = b^2 - 4ac

(c)

We have,

(k + 1)/k + (k + 2)/(k + 1) = -b/a (1)

and (k + 1)/k + (k + 2)/(k + 1) = c/a

⇒ (k + 2)/k = c/a or 2/k = c/a - 1 = (c - a)/a or k = 2a/(c - a) (2)

Now, eliminate k. Putting the value of k in Eq. (1), we get

(c + a)/2a + 2c/(c + a) = -b/a

⇒(c + a )^2 + 4ac = - 2b( a + c)

⇒( a + c )^2 + 2b( a + c) = -4ac

Adding b^2 to both sides, we have

(a + b + c )^2 = b^2 - 4ac

**Q2.**Let r,s and t be the roots of the equation, 8x^3 + 1001x + 2008 = 0. The value of (r+s)^3 + (s+t)^3 + (t+r)^3 is

Solution

(d)

Equation 8x^3 + 1001x + 2008 = 0 has roots r,s and t

r + s + r = 0,rst = -2008/8 = -251

Now, let r + s = A, s + 1 = B, t + r = C,

∴ A + B + C = 2 (r + s + t) = 0

Hence,

A^3 + B^3 + C^3 = 3ABC

∴ (r + s)^3 + ( s + t)^3 + ( t + r)^3

=3(r + s )( s + t)( t + r)

=3(r + s + t - t)(s + t + r - r)(t + r + s - s)

=-3rst(as r + s + t = 0 )

=3(251) = 753

(d)

Equation 8x^3 + 1001x + 2008 = 0 has roots r,s and t

r + s + r = 0,rst = -2008/8 = -251

Now, let r + s = A, s + 1 = B, t + r = C,

∴ A + B + C = 2 (r + s + t) = 0

Hence,

A^3 + B^3 + C^3 = 3ABC

∴ (r + s)^3 + ( s + t)^3 + ( t + r)^3

=3(r + s )( s + t)( t + r)

=3(r + s + t - t)(s + t + r - r)(t + r + s - s)

=-3rst(as r + s + t = 0 )

=3(251) = 753

**Q3.**If b>a , then the equation (x-a)(x-b) -1 = 0 has

**Q4.**If l,m,n are real l≠m, then the roots of the equation (l-m)x^2 - 5(l+m)x- 2(l-,m)=0 are

Solution

(c)

l , m , n are real and l≠m. Given equation is

(l - m ) x^2 - 5(l + m )x - 2( l - m) = 0

D = 25( l + m)^2 + 8( l - m)^2 > 0,l,m∈R

Therefore, the roots are real and unequal

.

(c)

l , m , n are real and l≠m. Given equation is

(l - m ) x^2 - 5(l + m )x - 2( l - m) = 0

D = 25( l + m)^2 + 8( l - m)^2 > 0,l,m∈R

Therefore, the roots are real and unequal

.

**Q5.**If the expression x^2 + 2(a+b+c)x + 3(bc+ca+ab) is a perfect square, then

Solution

(a)

Given quadratic expression is x^2+2(a+b+c)x+3(bc+ca+ab), this quadratic expression will be a perfect square if the discriminant of its corresponding equation is zero. Hence,

4(a + b + c)^2 - 4×3(bc + ca + ab) = 0

⇒(a + b + c)^2 - 3( bc + ca + ab) = 0

⇒a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3(bc + ca + ab ) = 0

⇒a^2 + b^2 + c^2 - ab - bc - ca = 0

⇒1/2 [ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ] = 0

⇒1/2 [(a^2 + b^2 - 2ab ) + ( b^2 + c^2 - 2bc ) + ( c^2 + a^2 - 2ca )] = 0

⇒1/2 [(a - b)^2 + (b -c )^2 + (c - a)^2 ] = 0

Which is possible only when

(a - b)^2 = 0,(b - c )^2 = 0 and ( c - a)^2 = 0, i.e., a = b = c

(a)

Given quadratic expression is x^2+2(a+b+c)x+3(bc+ca+ab), this quadratic expression will be a perfect square if the discriminant of its corresponding equation is zero. Hence,

4(a + b + c)^2 - 4×3(bc + ca + ab) = 0

⇒(a + b + c)^2 - 3( bc + ca + ab) = 0

⇒a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3(bc + ca + ab ) = 0

⇒a^2 + b^2 + c^2 - ab - bc - ca = 0

⇒1/2 [ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ] = 0

⇒1/2 [(a^2 + b^2 - 2ab ) + ( b^2 + c^2 - 2bc ) + ( c^2 + a^2 - 2ca )] = 0

⇒1/2 [(a - b)^2 + (b -c )^2 + (c - a)^2 ] = 0

Which is possible only when

(a - b)^2 = 0,(b - c )^2 = 0 and ( c - a)^2 = 0, i.e., a = b = c

**Q6.**If |z| < √2 - 1 , then |z^2 + 2z cosÎ± |is

Solution

(a)

|z^2 + 2z cosÎ± | ≤ | z^2 | + | 2z cosÎ± |

= |z|^2 + 2 | z || cosÎ± |

≤ |z|^2 + 2 | z |

< ( √2 - 1)^2 + 2(√2 - 1) = 1

(a)

|z^2 + 2z cosÎ± | ≤ | z^2 | + | 2z cosÎ± |

= |z|^2 + 2 | z || cosÎ± |

≤ |z|^2 + 2 | z |

< ( √2 - 1)^2 + 2(√2 - 1) = 1

**Q8.**If a,b∈R,a≠0 and the quadratic equation ax^2 -bx + 1 = 0 has imaginary roots then ( a + b + 1) is

Solution

(a) D = b^2 - 4a < 0 ⇒ a > 0

Therefore the graph is concave upwards

f(x) > 0 , ∀ x ∈R

⇒f(-1) > 0

⇒a + b + 1 > 0

(a) D = b^2 - 4a < 0 ⇒ a > 0

Therefore the graph is concave upwards

f(x) > 0 , ∀ x ∈R

⇒f(-1) > 0

⇒a + b + 1 > 0

**Q9.**Sum of the non-real roots of (x^2 + x - 2) ( x^2+ x - 3) = 12 is

Solution

(a)

Put x^2 + x = y, so that Eq.(1) becomes

(y - 2)(y - 3) = 12

⇒y^2 - 5y - 6 = 0

⇒(y - 6) ( y + 1)=0

⇒y = 6, -1

When y = 6, we get

x^2 + x - 6 = 0

⇒(x + 3)( x - 2) = 0 or x = -3 , 2

When y = -1, we get

Which has non-real roots and sum of roots is -1

(a)

Put x^2 + x = y, so that Eq.(1) becomes

(y - 2)(y - 3) = 12

⇒y^2 - 5y - 6 = 0

⇒(y - 6) ( y + 1)=0

⇒y = 6, -1

When y = 6, we get

x^2 + x - 6 = 0

⇒(x + 3)( x - 2) = 0 or x = -3 , 2

When y = -1, we get

Which has non-real roots and sum of roots is -1