Quadratic Equation

1 Let a, b, c ∈ R, a 6= 0. Then equation Q(x) = ax2 + bx + c = 0 is called

a real quadratic equation. ∆ = b2 − 4ac is called the discriminant of this
equation. A number α is called the root of Q(x) = 0 iff aα2 + bα + c = 0.
2 Nature of the roots of Q(x) = 0.
Roots are real and distinct iff ∆ > 0.
Roots are real and coincident iff ∆ = 0.
Roots are complex and conjugate iff ∆ < 0.
√ √
3 (i) If ∆ ≥ 0 then roots are given by α = −b−2a ∆ , β = −b+2a ∆
. Note that β
ingeneral is not greater than α. √ √
(ii) If ∆ < 0 then roots are given by −b−i2a −∆ , −b+i2a −∆ .
b2 −2ac
4 (i) α + β = − ab (ii) αβ = c
a
(iii) (α − β)2 = − a∆2 (iv) α2 + β 2 = a2
3
(v) α3 + β 3 = − b −3abc
a3
.
5 Transformation of Equation
Let roots of Q(x) = ax2 + bx + c = 0 be α, β.
(i) Equation whose roots are −α, − β is given by Q(−x) = 0.
(ii) Equation whose roots are α1 , β1 is given by x2 Q( x1 ) = cx2 + bx + a = 0.
(iii) Equation whose roots are α + k, β + k is given by Q(x − k) = 0.
(iv) Equation whose roots are f (α), f (β) is given by Q(f −1 (x)) = 0.
(v) Equation whose roots are α2 , β 2 is given by a2 x2 − (b2 − 2ac)x + c2 = 0.
(vi) Equation whose roots are α3 , β 3 is given by a3 x2 +(b3 −3abc)x+c3 = 0.
6 The graph of the function given by y = ax2 + bx + c, a 6= 0 is a Parabola.
∆ ∆
(i) Range of the function is [− 4a , +∞) if a > 0 and (−∞, − 4a ] if a < 0.
b ∆
(ii) The vertex of the parabola is (− 2a , − 4a ).
b
(iii) The graph is symmetric about the line x = − 2a .
(iv) The graph is above the X− axis iff a > 0 and ∆ < 0.
(v) The graph is below the X− axis iff a < 0 and ∆ < 0.
(vi) The graph touches the X− axis iff ∆ = 0.
(vii) The graph intersects the X− axis in two distinct points iff ∆ > 0.
7 The N&S condition that a1 x2 + b1 x + c1 = 0 and a2 x2 + b2 x + c2 = 0 have
2
at least one common root is Da,b Db,c = Da,c . To prove this (M1) Eliminate
x from given equations. (M2) Evaluate (α1 − α2 )(α1 − β2 )(β1 − α2 )(β1 − β2 )
and convert in terms of given coefficients.
8 The N&S condition that a1 x2 + b1 x + c1 = 0 and a2 x2 + b2 x + c2 = 0
have exactly same roots is aa12 = bb12 = cc12 . In simple words one equation is a
nonzero multiple of other.
9 Consider the equation Q(x) = ax2 + bx + c = 0, a 6= 0. Let a, b, c ∈ Q.
The roots are rational
√ iff ∆ = b2 − 4ac is square of a √
rational number. √
Further if u + v is a root with u, v ∈ Q, v > 0 and v ∈ Q′ then u − v
is also a root.
10 Possible Rational Root Thm Consider a polynomial equation of
degree n over Z, given by an xn + an−1 xn−1 + · · · + a0 = 0. One need to test
only those rational numbers of the type ± hk where h is a factor of a0 and k
is a factor of an for the possible rational root of the given equation.
Further if an = 1 then if given equation has a rational root then it must be
an integer.
11 Consider a polynomial equation an xn + an−1 xn−1 + · · · + a0 = 0 of degree
n over C. This equation has at most n distinct roots.
12 Consider a polynomial equation an xn + an−1 xn−1 + · · · + a0 = 0 over C.
If it has n + 1 distinct roots then the polynomial is identically zero.
13 Consider a polynomial equation of degree n over C, given by
P (x) = an xn + an−1 xn−1 + · · · + a0 = 0. Let α1 , α2 , · · · , αn denote its roots.
Then P (x) = an (x − α1 )(x − α2 ) · · · (x − αn ).
Let σi denote the sum of the roots taken i at a time then σi = aan−i n
(−1)n−i .
14 Let Si = α1i + α2i + · · · + αni . We have (i) S1 = σ1 (ii) S2 = σ12 − 2σ2
(iii) S3 = σ13 − 3σ1 σ2 + 3σ3 .
15 Let P (x) ∈ R[x]. If S2 < 0 then given equation can not have all real
roots.
16 Let P (x) ∈ R[x]. If degree of P (x) is odd then it must have at least one
real root. More over if α + iβ, β 6= 0 is a root of P (x) = 0 of multiplicity
m then α − iβ is also a root of multiplicity m.
17 Let P (x) ∈ R[x]. If P (a)P (b) < 0, a, b ∈ R then there exist odd number
of roots of P (x) = 0 between a, b. Note that some roots could be multiple
roots. Hence one can not say that there are odd distinct roots between a, b.
18 Descartes Rule of Signs Consider the real Degree n polynomial equa-
tion an xn + an−1 xn−1 + · · · + a1 x + a0 = 0. Scan the coefficients from right
to left and count the number of changes of signs. If these are k then the
number of real positive roots of given equation is one of the non negative
integers k, k − 2, k − 4, · · · .
19 To estimate the number of negative real roots replace x by −x and apply
the above method.
20 Graph of Real Cubic Consider y = P (x) = ax3 + bx2 + cx + d. Now
P ′ (x) = 3ax2 + 2bx + c. Three cases arise.
(i) P ′ (x) = 0 has no real roots (4b2 − 12ac ≤ 0) then P (x) is strictly
increasing (Decreasing) if a is positive (negative). Hence P (x) = 0 has only
one real root.
(ii)P ′ (x) = 0 has real coincident roots (4b2 − 12ac = 0) say α then P (x) is
strictly increasing (Decreasing) if a is positive (negative). Now P (x) = 0
has one or three coincident roots according as P (α) is non zero or is zero.
(iii) P ′ (x) = 0 has real distict roots. Now P (x) = 0 has one real root or two
of the real roots coincide or three distinct real roots according as P ′ (α)P ′ (β)
is +ve, zero or −ve. Further if a > 0 then P (x) has local maxima at α and
local minima at β. If a < 0 then then P (x) has local minima at α and local
b
maxima at β. Finally P (x) has unique point of inflexion at x = − 3a where
P (x) = 0
′′

21 Cardan’s Method WLG we start with the reduced cubic x3 +px+q = 0.

Step(i) Flexibility. Put x = u+v. We get (u3 +v 3 +q)+(u+v)(3uv +p) = 0.
Step(ii) Now solve (u3 + v 3 = −q) and uv = − p3 .
p3
Step(iii) Symmetrise. Actually solve (u3 + v 3 = −q) and u3 v 3 = − 27 .
2 p3
Step(iv) Leads to quadratic t + qt − 27 = 0.
Step(v) Threerof the foollwing nine
r numbers are the required roots.
q q
2 3 2 3
− 2q + q4 + 27
3 p
+ 3 − 2q − q4 + 27p

Nature of the roots is real distinct, (α, α, −2α) or only one real according
2 p3
as ∆ = q4 + 27 is −ve, zero or +ve.
22 Ferrari’s method for Quartic We write the quartic in the form
x4 + 2ax3 = bx2 + cx + d.
We add suitable quantities on RHS and LHS to obtain
(x2 + ax + λ)2 = bx2 + cx + d + 2λx2 + 2λax + λ2
Now we find λ such that RHS is a perfect square. For that ∆x = 0 which
turns out to be a cubic in λ hence has a real solution. For this λ the given
equation now factorises in two quadratic equations which can be solved.
23 Equations Reducible to Quadratic
(i) (x + a)(x + b)(x + c)(x + d) + k = 0 can be reduced provided sum of
the two of a, b, c, d equals the sum of the remaining two. For example if
a + b = c + d then put x2 + (a + b)x = y.
(ii) (x + a)(x + b)(x + c)(x + d) + kx2 = 0 can be reduced provided product
of the two of a, b, c, d equals the product of the remaining two. For example
if ab = cd then first divide by x2 and put x + ab x
= y.
(iii) In (x + a)4 + (x + b)4 = k put x + a+b2
= y.
(iv) Symmetric Quartic ax4 + bx3 + cx2 + bx + a = 0. First regroup as
x2 a[(x + x1 )2 − 2] + b(x + x1 ) + c = 0 and now put y = (x + x1 ).
4 3 2
(iv) Almost
 Symmetric Quartic ax  +bx +cx −bx+a = 0. First regroup
as x2 a[(x − x1 )2 + 2] + b(x − x1 ) + c = 0 and now put y = (x − x1 ).
24 Consider the quadratic equation ax2 + bx + c = 0.
The coefficients a, b, c depend upon parameter m.
Let α, β denote the roots of the given quadratic equation.
If the roots are real, we will assume that α ≤ β.
P R denotes the product of the roots. SR denotes the sum of the roots.
k, k1 , k2 ∈ R. Let ∆ = b2 − 4ac and f (x) = ax2 + bx + c.
It is assumed that a 6= 0.
If a = 0, then corresponding cases must be considered separately.
The following table gives the necessary and sufficient crieterions.

Conditions imposed restrictions

α, β > 0 (1) ∆ ≥ 0 and (2) SR > 0 and (3) P R > 0.
α, β ≥ 0 (1) ∆ ≥ 0 and (2) SR ≥ 0 and (3) P R ≥ 0.
α, β < 0 (1) ∆ ≥ 0 and (2) SR < 0 and (3) P R > 0.
α, β ≤ 0 (1) ∆ ≥ 0 and (2) SR ≤ 0 and (3) P R ≥ 0.
At least one (1) ∆ ≥ 0 and
root +ve (2a) SR > 0 and P R ≥ 0 or (2b) P R < 0.
At least one (1) ∆ ≥ 0 and
root −ve (2a) SR < 0 and P R ≥ 0 or (2b) P R < 0.
α<k<β af (k) < 0.
k1 < α ≤ β (1) ∆ ≥ 0 and (2) af (k1 ) > 0 and (3) −b/2a > k1 .
k1 ≤ α ≤ β (1) ∆ ≥ 0 and (2) af (k1 ) ≥ 0 and (3) −b/2a ≥ k1 .
α ≤ β < k2 (1) ∆ ≥ 0 and (2) af (k2 ) > 0 and (3) −b/2a < k2 .
α ≤ β ≤ k2 (1) ∆ ≥ 0 and (2) af (k2 ) ≥ 0 and (3) −b/2a ≤ k2 .
α < k1 ≤ k2 < β (1) af (k1 ) < 0 and (2) af (k2 ) < 0.
α ≤ k1 < k2 ≤ β (1) af (k1 ) ≤ 0 and (2) af (k2 ) ≤ 0.
The correct way to remember this table is to visualise the graph of the
quadratic equation and place the position of the vertex correctly.
25 The N&S condition for ax4 + bx2 + c = 0 to have real roots is
(1) ∆ ≥ 0 and (2a) (SR ≥ 0 and P R ≥ 0) or (2b) P R ≤ 0.