Question #2a730

3 Answers
Sep 10, 2016

Answer:

#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))#

Sep 10, 2016

Answer:

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)#.

Sep 10, 2016

Answer:

#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.!