**Illustration 1.** Prove that the roots of ax^{2} + 2bx + c = 0 will be real and distinct if and only if the roots of (a + c) (ax^{2} + 2bx + c) = 2(ac − b^{2}) (x^{2} + 1) are imaginary.

**Solution:** Let D_{1} be the discriminant of ax^{2} + 2bx + c = 0 … (1)

The other equation (a + c) (ax^{2} + 2bx + c) = 2(ac − b^{2}) (x^{2} +1) can be written as

(a^{2} + 2b^{2} − ac ) x^{2} + 2(ab + bc) x + (c^{2} + 2b^{2} − ac) = 0 ….(2)

Hence D_{1} = 4b^{2} − 4ac = 4 (b^{2} − ac) and

D_{2} = 4(ab + bc)^{2} − 4( c^{2} + 2b^{2} − ac) (a^{2} +2b^{2} − ac),

where D_{2} is the discriminant of equation (2)

= 4 (− a^{2}b^{2} − b^{2} c^{2} − 2a^{2}c^{2} + 6ab^{2}c − 4b^{4} + ac^{3} + a^{3}c)

= − 4[ 4b^{2}(b^{2} − ac) + a^{2}(b^{2} − ac)+ c^{2}(b^{2} − ac) − 2ac(b^{2} − ac) ]

= − 4(b^{2} − ac)(4b^{2} + (a − c)^{2})

Since D_{2} < 0 <=> b^{2} − ac > 0

or, D_{2} < 0 <=> D_{1} > 0,

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

**Illustration 2 :** Let a and b be the roots of the equation ax^{2} + 2bx + c = 0 and a + g and b + g be the roots of Ax^{2} + 2Bx + C = 0. Then prove that A^{2}(b^{2} − ac) = a^{2}(B^{2} − 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} − 4α β

=> a^{2} (B^{2} − AC) = A^{2}(b^{2} − ac)