# Problem 1. If the roots of ax^2 + bx + c = 0 are 2 and 8 and the roots of bx^2 + cx + a = 0 are 1 and 2, what are a, b and c.

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### 3 Answers

The roots of ax^2 + bx + c = 0 are 2 and 8. As the roots of the equation are 2 and 8, it can be written in the following way:

(x - 2)(x - 8) = ax^2 + bx + c

=> x^2 - 10x + 16 = ax^2 + bx + c

=> b/a = -10 and c/a = 16

The roots of bx^2 + cx + a = 0 are 1 and 2. This can be written as (x - 1)(x - 2) = bx^2 + cx + a

=> x^2 - 3x + 2 = bx^2 + cx + a

=> c/b = -3 and a/b = 2

The equations give b/a = -10 and a/b = 2. This is not possible as `2 != -1/10` .

There are no values of a, b, c such that the two equations have the given roots.

Here is another method of solving this problem.

We know that a quadratic equation of the form ax^2 + bx + c = 0 has two solutions:

[- b + sqrt(b^2 - 4ac)] / 2a and [- b - sqrt(b^2 - 4ac)] / 2a

So for ax^2 + bx + c = 0 with solutions x = 2 and x= 8, we have:

[- b + sqrt(b^2 - 4ac)] / 2a = 2

or [- b + sqrt(b^2 - 4ac)] = 4a .... (1)

and

[- b - sqrt(b^2 - 4ac)] / 2a = 8

or [- b - sqrt(b^2 - 4ac)] = 16a .... (2)

Multiplying (1) by (2), we get:

[- b + sqrt(b^2 - 4ac)] * [- b - sqrt(b^2 - 4ac)] = 4a * 16a

or (-b)^2 - [sqrt(b^2 - 4ac)]^2 = 64a^2 as (X + Y) (X - Y) = X^2 - Y^2

or b^2 - (b^2 - 4ac) = 64a^2

or b^2 - b^2 + 4ac = 64a^2

or 4ac = 64a^2

or c = 16a

Similarly, for the second quadratic equation:

[- c + sqrt(c^2 - 4ba)] / 2b = 1

or [- c + sqrt(c^2 - 4ba)] = 2b ..... (3)

and

[- c - sqrt(c^2 - 4ba)] / 2b = 2

or [- c - sqrt(c^2 - 4ba)] = 4b .... (4)

So similarly multiplying (3) by (4), we get:

[- c + sqrt(c^2 - 4ba)] * [- c - sqrt(c^2 - 4ba)] = 2b * 4b

or (-c)^2 - [sqrt(c^2 - 4ba)]^2 = 8b^2

or c^2 - (c^2 - 4ba) = 8b^2

or 4ba = 8b^2

or a = 2b

Finally we have: c = 16a and a = 2b

So, a : b : c = 2b : b : 16a = 2b : b : 32b = 2 : 1 : 32

Considering smallest integral ratio,

a = 2, b = 1 and c = 32

If we plug in the values of a, b, and c as well as the x-values to the quadratic equation, we arrive at a 'mathematical fallacy' where we get a non-zero quantity equals to zero. Hence, I guess, the a, b, and c values are not solvable.

[Note: The question implies that the values of a, b and c will hold true for both quadratic equations, so treating each quadratic equation separately and finding a set of a, b, and c values for each equation, is not the way to go.]

we know for any quadratic equation ax^2+bx+c=0 sum of roots is -b/a and product of roots is c/a.``

so for first equation we get c/a=16

for second equation we get-c/a=3 i.e. c/a=-3.But 16`!=` 3 hence not possible.``

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We do not get : -c/a = 3 from the second equation. So no question of comparing it with 16. This is a wrong answer!