Basic Concepts: An equation of the form ax2 + bx + c = 0, where a ≠0 and a, b, c are real numbers, is called a quadratic equation.

The numbers a, b, c are called the coefficients of the quadratic equation.

A root of the quadratic equation is a number α (real or complex) such that aα2 + bα + c = 0.

The roots of the quadratic equation are given by

$\displaystyle x = \frac{-b \pm \sqrt{b^2 – 4a c}}{2 a}$

The quantity D (D = b2 – 4ac) is known as the discriminant of the equation

Basic Results:

(i) The quadratic equation has real and equal roots if and only if D = 0

i.e. b2 – 4ac = 0.

(ii) The quadratic equation has real and distinct roots if and only if D > 0

i.e. b2 – 4ac > 0.

(iii) The quadratic equation has complex roots with non-zero imaginary parts if and only if D < 0

i.e. b2 – 4ac < 0.

(iv) If p + iq (p and q being real) is a root of the quadratic equation where i = √-1 , then p – iq is also a root of the quadratic equation.

(v) If p + √q is an irrational root of the quadratic equation, then p – √q is also a root of the quadratic equation provided that all the coefficients are rational , q not being a perfect square or zero.

(vi) The quadratic equation has rational roots if D is a perfect square and a, b, c are rational.

(vii) If a = 1 and b, c are integers and the roots of the quadratic equation are rational, then the roots must be integers.

(viii) If the quadratic equation is satisfied by more than two distinct numbers (real or complex), then it becomes an identity i.e. a = b = c = 0.

(ix) Let α and β be two roots of the given quadratic equation. Then α + β = -b/a and αβ = c/a .

(x) A quadratic equation, whose roots are α and β can be written as

(x – α) (x – β) = 0

i.e., ax2 + bx + c ≡ a(x – α) (x – β).

Illustration 1. Prove that the roots of ax2 + 2bx + c = 0 will be real and distinct if and only if the roots of (a + c) (ax2 + 2bx + c) = 2(ac − b2) (x2 + 1) are imaginary.

Solution: Let D1 be the discriminant of

ax2 + 2bx + c = 0 … (i)

The other equation (a + c) (ax2 + 2bx + c) = 2(ac − b2) (x2 +1) can be written as

(a2 + 2b2 − ac ) x2 + 2(ab + bc) x + (c2 + 2b2 − ac) = 0 ….(ii)

Hence , D1 = 4b2 − 4ac

D1 = 4 (b2 − ac)

and D2 = 4(ab + bc)2 − 4( c2 + 2b2 − ac) (a2 +2b2 − ac),

where D2 is the discriminant of equation (ii)

= 4 (− a2b2 − b2 c2 − 2a2c2 + 6ab2c − 4b4 + ac3 + a3c)

= − 4[ 4b2(b2 − ac) + a2(b2 − ac)+ c2(b2 − ac) − 2ac(b2 − ac) ]

= − 4(b2 − ac)(4b2 + (a − c)2)

Since D2 < 0

⇒ b2 − ac > 0

or, D2 < 0

=> D1 > 0

the roots of (1) are real and distinct if and only if the roots of (2) are imaginary.

Illustration 2 : Let α and β be the roots of the equation ax2 + 2bx + c = 0 and α + γ and β + γ be the roots of Ax2 + 2Bx + C = 0. Then prove that A2(b2 − ac) = a2(B2 − AC).

Solution: For the given equation, α + β = − 2b/a , α β = c/a

and α + β + 2γ = -2B / A , (α + γ) (β + γ) = C / A

((α + γ) – β + γ)) 2 = ( α + β) 2 – 4 α β

$\displaystyle \frac{4B^2}{A^2} – \frac{4C}{A} =\frac{4b^2}{a^2} – \frac{4c}{a}$

⇒ a2 (B2 − AC) = A2(b2 − ac)

Illustration 3: Let f (x) = Ax2 + Bx + C where A, B, C are real numbers. Prove that if , f (x) is an integer whenever x is an integer, then the numbers 2A , A + B and C are all integers. Conversely, prove that if the numbers 2A, A + B and C are all integer then f (x) is an integer whenever x is an integer.

Solution: Let us consider the integral values of x as 0 , 1 , − 1 then f (0) , f (1) and f (- 1) are all integers

=> f (0) = C , f (1) = A + B + C and f (- 1) = A – B + C all integers

Therefore C is an integer and hence A + B is an integer also A − B is an integer

Now, 2A = (A + B) + (A – B) = sum of two integers

=> 2A is also an integer

=> 2A , A + B and C are all integers

conversely let n ∈ I then f (n) = An2 + Bn + C

= 2A[n(n − 1)/2] + (A + B) n + C

now n(n − 1)/2 = even/2 = integer and 2A , A + B and C are also integer

=> f (x) is an integer

Exercise :

(i) If sinθ, cosθ are the roots of the equation ax2 + bx + c = 0 then find the value of

$\displaystyle \frac{(a+c)^2}{b^2 + c^2}$

(ii) If the sum of the roots of the equation ax2 + bx + c = 0 is equal to sum of the squares of their reciprocals, then show that ab2 , a2c , bc2 are in A.P.

(iii) If the roots α and β of the quadratic equation ax2 + bx + c = 0 are real and of opposite sign. Then show that roots of the equation α(x − β)2 + β(x − α)2 = 0 are also real and of opposite sign.