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

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

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.

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.

14 comments:

  1. Replies
    1. Thank you so much for such an easy solution!

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    2. exactuly

      thanks for show easy solution

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  2. Tank you so much!

    ReplyDelete
  3. its very long can you suggest a smaller method

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  4. Sir the answer is of great standard thanks

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  5. Amazing logic.Thanks a lot.

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  6. This comment has been removed by the author.

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  7. Hi, I found a solution with few steps.

    ax²+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 ^_~

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