(a-b)x2+(b-c)z+(c-a)=0
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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.
:)
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.
:)
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