If the roots of the equation (b - c) x square (c-a) x (a-b) =0 are equal, prove that 2b =a + c. Its urgent help me?

Question

Answer

Given that the quadratic equation,
(b-c)x²+(c-a)x+(a-b)=0,
has two equal roots,
then discriminant=0,
b²-4ac=0,
(c-a)²-4(b-c)(a-b)=0,
c²+a²-2ac-(4(ab-b²-ca+bc))=0,
c²+a²-2ac-4ab+4b²+4ac-4bc=0,
c²+a²+2ac-4ab+4b²-4bc=0
on further simplification,
2b=a+c

Answer

Your answer is here...

(b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

A=(b-c),
B=c-a,
C=a-b

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b

Answer

Answer

Understanding the Equation
The given equation is
(b - c) x^2 + (c - a)x + (a - b) = 0.
To analyze when the roots are equal, we need to focus on the discriminant of a quadratic equation.
Discriminant Condition
For a quadratic equation ax^2 + bx + c = 0, the roots are equal when the discriminant (D) is zero. The discriminant is given by:
D = b^2 - 4ac.
In our case,
- a = (b - c)
- b = (c - a)
- c = (a - b)
Thus, the discriminant becomes:
D = (c - a)^2 - 4(b - c)(a - b).
Setting the Discriminant to Zero
To find when the roots are equal:
(c - a)^2 - 4(b - c)(a - b) = 0.
Expanding the Expression
1. Expand (c - a)^2:
- (c - a)^2 = c^2 - 2ac + a^2.
2. Expand 4(b - c)(a - b):
- 4(b - c)(a - b) = 4(ba - b^2 - ac + bc).
3. Combine these results in the equation.
Final Condition
After simplification, you will arrive at:
2b = a + c.
This shows that when the roots of the equation are equal, it leads to the condition that 2b = a + c, proving the required statement.
Conclusion
Thus, we have demonstrated through the discriminant condition and simplification that for the roots of the given quadratic equation to be equal, the relationship 2b = a + c must hold true.

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