PROVE THAT PROBLEMS USING THE CONCEPT NATURE OF QUADRATIC EQUATION

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 + b³  =  3abc

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