Question #2a730

3 Answers
Sep 10, 2016

x = (- b + c pm sqrt((b + c)^(2) + 4 a (a - b - c))) / (2 (a - b))x=b+c±(b+c)2+4a(abc)2(ab)

Explanation:

We have: (a - b) x^(2) + (b - c) x + (c - a) = 0(ab)x2+(bc)x+(ca)=0

Let's apply the quadratic formula:

x = (- (b - c) pm sqrt((b - c)^(2) - (4) (a - b) (c - a))) / (2 (a - b))x=(bc)±(bc)2(4)(ab)(ca)2(ab)

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±b22bc+c2(4)(aca2bc+ab)2(ab)

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±b22bc+c24ac+4a2+4bc4ab2(ab)

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±4a2+b2+c24ab+2bc4ac2(ab)

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±(b2+2bc+c2)+(4a24ab4ac)2(ab)

x = (- b + c pm sqrt((b + c)^(2) + 4 a (a - b - c))) / (2 (a - b))x=b+c±(b+c)2+4a(abc)2(ab)

Sep 10, 2016

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

Explanation:

Let p(x)=(a-b)x^2+(b-c)x+(c-a)=0p(x)=(ab)x2+(bc)x+(ca)=0.

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

p(x)=(a-b)+(b-c)+(c-a)=0p(x)=(ab)+(bc)+(ca)=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).

Sep 10, 2016

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.!