if ax2+bx+c=0and bx2+cx+a=0 have a common root ,then prove that a+b+c=0 or a=b=c. (it's an advanced maths question plse answer correctly i need it)......help me
Answers
Step-by-step explanation:
Since it is a common root, it should satisfy both these equations. So, we'll have
at^2 + bt + c = 0
bt^2 + ct + a = 0
By substracting the second equation from the first,
at^2 + bt + c - bt^2 - ct - a = 0
Now, we'll pair the terms as indicated by the colours
=> at^2 - a - bt^2 + bt - ct + c = 0
=> a(t^2 - 1) - bt(t - 1) - c(t - 1) = 0
We can take (t - 1) common from the expression, because t^2 - 1 = (t + 1)(t - 1)
=> (t - 1) [ a(t + 1) - bt - c] = 0
=> t - 1 = 0, or a(t + 1) - bt - c =0
From t - 1 = 0, we get t = 1. This gives us that the common root is 1 and substituting it in either of the equation will give a + b + c = 0
We'll go on with a(t + 1) - bt - c = 0
=>at - bt = c - a
=> t = (c - a)/(a - b)
By substituting this value of t in at^2 + bt + c = 0, we'll get
ax^2 + bx + c = 0 and bx^2 + cx + a = 0 have a common root. Prove that a+b+c=0 or a=b=c
=> a = 0, (a - b)^2 + (b - c)^2 + (c - a)^2 = 0
Now, (a - b)^2, (b- c)^2 and (c - a)^2 are all squares, so they'll be greater than or equal to zero. The expression, which is their sum, can be 0 only in the case when each of them is equal to 0.
=> (a - b)^2 = 0, (b - c)^2 = 0, (c - a)^2 = 0
=> a = b, b = c, c = a
or a = b = c
Hence Proved.
Answer:
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