HOTS : Quadratic Equation

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 … (1)

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 ….(2)

Hence D1 = 4b2 − 4ac = 4 (b2 − ac) and

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

where D2 is the discriminant of equation (2)

= 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 a and b be the roots of the equation ax2 + 2bx + c = 0 and a + g and b + g 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 − 4α β

=> a2 (B2 − AC) = A2(b2 − ac)

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