We are given the equations ax^2 + bx + c = 0 and bx^2 + cx + a = 0 and we are told that they have a common root. Lets assume the common root to be t.
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
=> 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.
NOTE - We must also understand the significance of the question and the solution. We began with the assumption of a common root between the two equations. Furthur operations with the two resultant equations gave us 3 results: a + b + c = 0, a = 0, a = b = c. This actually means that if even one of these three results holds true, the equations will have a common root.
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
=> 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.
NOTE - We must also understand the significance of the question and the solution. We began with the assumption of a common root between the two equations. Furthur operations with the two resultant equations gave us 3 results: a + b + c = 0, a = 0, a = b = c. This actually means that if even one of these three results holds true, the equations will have a common root.
THANX
ReplyDeleteThank you so much for such an easy solution!
Deleteexactuly
Deletethanks for show easy solution
Tank you so much!
ReplyDeletethanxðŸ˜ðŸ˜
ReplyDeleteits very long can you suggest a smaller method
ReplyDeleteSir the answer is of great standard thanks
ReplyDeleteThanku
ReplyDeleteThank a lot.
ReplyDeleteAmazing logic.Thanks a lot.
ReplyDeleteNo problem :)
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ReplyDeleteHi, I found a solution with few steps.
ReplyDeleteax²+bx+c=0 ← 1
bx²+cx+a=0 ← 2
If equation 1 and 2 common root, then there coefficients are proportional.
∴ (a/b)=(b/c)=(c/a)
Let's say the proportions are equal to k
∴ (a/b)=(b/c)=(c/a) = k
Now,
a=bk, b=ck, c=ak
{If we add these we get
a+b+c=k(a+b+c) ←we need this equation later
}
lets take the proportions (a/b) and (b/c)
we know both of them are equal to k
so,
(a/b)=k ← say equation x₁
(b/c)=k ← say equation x₂
Now multiply these two,
(x₁)(x₂) ⇒ (ab/bc)=k²
Simplify and u get,
= (a/c)=k²
= a=ck²
Now similarly, do the same for the other proportions. And u will get
a=ck² ← say equation R₁
b=ak² ← say equation R₂
c=bk² ← say equation R₃
Add R₁, R₂, R₃
R₁ + R₂ + R₃ ⇒ a+b+c=k²(a+b+c)
= k²=(a+b+c)/(a+b+c)
= k² = 1
∴ k=+1 or k=-1
When k=1
Apply k=1 for this equation (a/b)=(b/c)=(c/a) = k
∴ (a/b)=(b/c)=(c/a) = 1
∴ a=b , b=c , c=a
∴ a=b=c
When k=-1
Now u might remember the equation I told u to keep in mind. We derived it ealier
Apply k=-1 for a+b+c=k(a+b+c)
∴ a+b+c=-1(a+b+c)
= 2(a+b+c)=0
∴ a+b+c=0 ^_~
Hmm its easy
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