# PROVE THAT PROBLEMS USING THE CONCEPT NATURE OF QUADRATIC EQUATION

Prove that Problems Using the Concept Nature of Quadratic Equation :

Here we are going to see some example problems based on finding nature of quadratic equation.

 Value of discriminantΔ = b2 - 4acΔ > 0Δ = 0Δ < 0 Nature of rootsReal and unequal rootsReal and equal rootsNo real roots

## Prove that Problems Using the Concept Nature of Quadratic Equation - Questions

Question 1 :

If the roots of (a −b)x2 + (b −c)x + (c −a) = 0 are real and equal, then prove that b, a, c are in arithmetic progression

Solution :

By comparing the given quadratic equation with the general form of quadratic equation.

ax2 + bx + c = 0

a = a - b, b = b - c and c = c - a

If the roots are real and equal, then Δ = 0

Δ = b2 - 4ac

(b -  c)2 - 4(a - b)(c - a)  =  0

b2 + c2 - 2bc  - 4(ac - a2 - bc + ab)  =  0

b2 + c2 - 2bc  - 4ac + 4a2 + 4bc - 4ab  =  0

b2 + c2 + 2bc  - 4ac + 4a2 - 4ab  =  0

4a2 + b2 + c2 - 4ab + 2bc  - 4ac  =  0

(2a)2 + (-b)2 + (-c)2 + 2(2a)(-b) + 2(-b)(-c)  + 2(2a)(-c) =  0

(2a - b - c)2  =  0

2a  =  b + c

Hence b, a and c are in A.P

Note :

If b, a and c are in A.P, then

a - b  =  c - a

2a  =  b + c

Question 2 :

If a, b are real then show that the roots of the equation

(a − b)x2 −6(a + b)x −9(a − b)  =  0 are real and unequal.

Solution :

a = a - b, b = -6(a + b) and c = -9(a - b)

Δ = b2 - 4ac

=  [-6(a + b)]2 - 4(a - b)(-9)(a - b)

=  36(a + b)2 + 36(a - b)2

=  36a2 + 72ab + 36b2 + 36a2- 72ab + 36b2

=  72(a2 + b2) > 0

Hence the roots are real and unequal.

Question 3 :

If the roots of the equation (c2 −ab)x2 −2(a2 −bc)x +b2 −ac = 0 are real and equal prove that either a = 0 (or) a3 +b3 +c3 = 3abc

Solution :

a = c2 −ab, b = −2(a2 −bc) and c = b2 −ac

Δ = b2 - 4ac

=  (−2(a2 −bc))2 - 4(c2 −ab)(b2 −ac)

=  4(a4 - 2a2bc  + b2c2) - 4(b2c2 - ac3 - ab3 + a2bc)

=  4a4 - 8a2bc  + 4b2c2 - 4b2c2 + 4ac3 + 4ab3 - 4 a2bc

=  4a4 - 12a2bc  + 4ac3 + 4ab3

4a(a3 - 3abc + c3 + b3)  =  0

a = 0 (or)  a3 + c3 + b=  3abc

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