Test: Quadratic Equation - 1 - JEE MCQ

Consider
For real roots,

Product of real roots

are in A.P. are in A.P.
…
are in A.P.
…
From & we get,
a = −b = c
⇒ α, β are roots of equation x² − x + 1 = 0

Since,

And

So, the required equation is


Sum of roots = 3

Given that α and β are the roots of the equation
x² + bx + c = 0
α + β = −b and αβ = c
Also α + h and β + h are the roots of the equations
We have,
x2 + qx  +r = 0
∴ α + h + β + h= −q
−b + 2h = −q

or

or

Hence,

Let α be the common root of given equations, then α² + bα − 1 = 0  and α² + α + b = 0
Subtracting (2) from (1), we get (b − 1)α − (b + 1) = 0 or α = b + 1 / b − 1
Substituting this value of α in equation (1), we get

or

Given: (ℓ − m)x² + ℓx + 1 = 0

Let roots be r, 2r

So, 

⇒2ℓ² − 9ℓ + 9m = 0
if ℓ is real, then D ⩾ 0
81 ⩾ 8 × 9 m

 Hence maximum value of m = 9/8

We have α² = 5α − 3 and β² = 5β − 3; ⇒ α & β are roots of equation, x² = 5x − 3 or x² − 5x + 3 = 0
∴ α + β = 5 and αβ = 3
Thus, the equation having α / β and β / α as its roots is

OR

Let and be the roots of
Now, and
Consider
Squaring both side,

and are the roots of or
and


Desired equation is
or

Given quadratic equation
is whose one root is
Consider

∴ Another root will be (∴ complex roots always occurs in pairs )
Thus, sum of roots = 

and product of roots = 

∴ Required equation is x² − ( sum of roots )x+( product of roots ) = 0

Thus, the values of a,b,c are 6,−4,1 respectively

Case 1: x ≥ 0
∴ the equation becomes or but
∴ both values, non admissible :
Case 2:  x ≤  0
The eq becomes or x = 1, 4 both values are non admissible,
∴ No real roots.
Alternatively, since
∴ x² + |x| + 4 > 0  for all x ∈ R
∴ x² + |x| + 4 > 0 for any x ∈ R

Consider both equations ...(i)
and ...(ii)
Since, both the equations are quadratic and have real roots, therefore from equation (1), we have
(using discriminant)
...(iii)
and from second equation
...(iv)
From eqs. (iii) and (iv) we get .

Therefore, f(x) = x² − 3x + 2 = 0 has roots 1 and 2. 
∴ α² + β² = 5

since

Since the roots are imaginary ∴D < 0 and roots occur as conjugate pair, i.e.

Also, let

Since az² + bz + c  =0 ...(1)

and z₁,z₂ (roots of (1)) are such that Im(z₁z₂) ≠ 0. Now, z₁ and z₂ are not conjugates of each other Complex roots of (1) are not conjugate of each other

Coefficient a,b,c cannot all be real at least one of a,b,c, is imaginary.

Let the discriminant of the equation be , then and
the discriminant of the equation is
[from the given relation]

Clearly at least one of D₁ and D₂ must be non-negative consequently at least one of the equation has real roots.