Question

Question #2a730

Answer 1

`x = (- b + c pm sqrt((b + c)^(2) + 4 a (a - b - c))) / (2 (a - b))`

Explanation:

We have: `(a - b) x^(2) + (b - c) x + (c - a) = 0`

Let's apply the quadratic formula:

`x = (- (b - c) pm sqrt((b - c)^(2) - (4) (a - b) (c - a))) / (2 (a - b))`

`x = (- b + c pm sqrt(b^(2) - 2 b c + c^(2) - (4) (a c - a^(2) - b c + a b))) / (2 (a - b))`

`x = (- b + c pm sqrt(b^(2) - 2 b c + c^(2) - 4 a c + 4 a^(2) + 4 b c - 4 a b)) / (2 (a - b))`

`x = (- b + c pm sqrt(4 a^(2) + b^(2) + c^(2) - 4 a b + 2 b c - 4 a c)) / (2 (a - b))`

`x = (- b + c pm sqrt((b^(2) + 2 b c + c^(2)) + (4 a^(2) - 4 a b - 4 a c))) / (2 (a - b))`

`x = (- b + c pm sqrt((b + c)^(2) + 4 a (a - b - c))) / (2 (a - b))`

Answer 2

The Roots are `1, and, (c-a)/(a-b)`.

Explanation:

Let `p(x)=(a-b)x^2+(b-c)x+(c-a)=0`.

We observe taht the sum of the co-effs. of

`p(x)=(a-b)+(b-c)+(c-a)=0`.

`:. (x-1)" is a factor of "`p(x)#.

Hence, one root of `p(x)=0" is 1."`

Next, to find the other root, we recall that,

The Product of roots of `Ax^2+Bx+C=0` ic `C/A`.

`":. in our case, the products of roots="(c-a)/(a-b)`, i.e.,

`(1)*(other root)=(c-a)/(a-b) rArr "the other root="(c-a)/(a-b)`.

Thus, the soln. is, `x=1, x=(c-a)/(a-b)`.

Answer 3

`x=1, x=(c-a)/(a-b)`.

Explanation:

Let `p(x)=lx^2+mx+n=0`, where,

`l=a-b, m=b-c, &, n=c-a`.

Note that, `l+m+n=(a-b)+(b-c)+(c-a)=0`, so,

writing `-l-n=m`, we have,

#p(x)=lx^2+(-l-n)x+n

`=ul(lx^2-lx)-ul(nx+n)`

`=lx(x-1)-n(x-1)`

`=(x-1)(lx-n)=0.`

Hnce, the roots are, `x=1, x=n/l=(c-a)/(a-b)`.

Enjoy Maths.!