If the roots of the equation
(b-c)x² (c-a)x (a-b)=0 are equal then prove that 2b=ac
Bipashanayak:
Actually q. Is little wrong
Answers
Answered by
2
Hi,
We know that,
If the quadratic equation ax²+bx+c=0
whose roots are equal then it's
deteminant is equal to zero.
(a-b)x²+(b-c)x+(c-a)=0
Deteminant =0
(b-c)² -4(a-b)(c-a)==0
b²+c²-2bc-4ac+4a²+4bc-4ab=0
b²+c²+4a²+4bc-4ac-4ab=0
b²+c²+(-2a)²+2bc+2c(-2a)+2(-2a)b=0
(b+c-2a)²=0
b+c-2a=0
Therefore,
b+c=2a
Hence proved.
I hope this helps you.
:)
We know that,
If the quadratic equation ax²+bx+c=0
whose roots are equal then it's
deteminant is equal to zero.
(a-b)x²+(b-c)x+(c-a)=0
Deteminant =0
(b-c)² -4(a-b)(c-a)==0
b²+c²-2bc-4ac+4a²+4bc-4ab=0
b²+c²+4a²+4bc-4ac-4ab=0
b²+c²+(-2a)²+2bc+2c(-2a)+2(-2a)b=0
(b+c-2a)²=0
b+c-2a=0
Therefore,
b+c=2a
Hence proved.
I hope this helps you.
:)
Answered by
1
SOLUTION :
Given : (b - c) x² + (c - a)x + (a - b) = 0 .
On comparing the given equation with ax² + bx + c = 0
Here, a = (b - c), b = (c - a), c = (a - b)
Discriminant(D) = b² -2ac
D = (c - a)² - 4(b - c)(a - b)
D = c² + a² - 2ac – 4(ab - b² - ac + cb)
[(a - b)² = a² + b² - 2ab]
D = c² + a² - 2ac – 4ab + 4b² + 4ac - 4cb
D = c² + a² - 2ac + 4ac – 4ab + 4b² - 4cb
D = c² + a² + 2ac – 4ab + 4b² - 4cb
D = a² + 4b² + c² + 2ac - 4ab - 4cb
As We know that , (a + c - 2b)² = a² + 4b² + c² + 2ac - 4ab - 4cb
D = (a + c - 2b)²
Real and equal root are Given
Discriminant(D) = 0
0 = (a + c - 2b)²
Put Square root on both sides
0 = √(a + c - 2b)²
0 = (a + c - 2b)
a + c = 2b
Hence , 2b = a + C
HOPE THIS ANSWER WILL HELP YOU….
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