if the roots of the quadratic equation (a-b)x² + (b-c)x + (c-a)=0 are equal prove their a²-b²+2ac =0
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Answer:
down there
Step-by-step explanation:
by putting these values in equation 1... we get...
(c-a)2 - 4 (b-c) (a-b) = 0
c2+a2-2ac - 4(ab-b2-ac+bc) = 0
c2+a2-2ac-4ab+4b2+4ac - 4bc = 0
a2 + 4b2 + c2 - 4ab - 4bc + 2ac = 0
(a-2b+c)2 = 0
so....
a -2b + c = 0
a + c = 2b
by putting these values in equation 1... we get...
(c-a)2 - 4 (b-c) (a-b) = 0
c2+a2-2ac - 4(ab-b2-ac+bc) = 0
c2+a2-2ac-4ab+4b2+4ac - 4bc = 0
a2 + 4b2 + c2 - 4ab - 4bc + 2ac = 0
(a-2b+c)2 = 0
so....
a -2b + c = 0
a + c = 2b
by putting these values in equation 1... we get...
(c-a)2 - 4 (b-c) (a-b) = 0
c2+a2-2ac - 4(ab-b2-ac+bc) = 0
c2+a2-2ac-4ab+4b2+4ac - 4bc = 0
a2 + 4b2 + c2 - 4ab - 4bc + 2ac = 0
(a-2b+c)2 = 0
so....
a -2b + c = 0
a + c = 2b
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