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 discriminant Δ = b2 - 4ac Δ > 0 Δ = 0 Δ < 0 |
Nature of roots Real and unequal roots Real and equal roots No real roots |
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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