# Question 2a730

Sep 10, 2016

$x = \frac{- b + c \pm \sqrt{{\left(b + c\right)}^{2} + 4 a \left(a - b - c\right)}}{2 \left(a - b\right)}$

#### Explanation:

We have: $\left(a - b\right) {x}^{2} + \left(b - c\right) x + \left(c - a\right) = 0$

$x = \frac{- \left(b - c\right) \pm \sqrt{{\left(b - c\right)}^{2} - \left(4\right) \left(a - b\right) \left(c - a\right)}}{2 \left(a - b\right)}$

$x = \frac{- b + c \pm \sqrt{{b}^{2} - 2 b c + {c}^{2} - \left(4\right) \left(a c - {a}^{2} - b c + a b\right)}}{2 \left(a - b\right)}$

$x = \frac{- b + c \pm \sqrt{{b}^{2} - 2 b c + {c}^{2} - 4 a c + 4 {a}^{2} + 4 b c - 4 a b}}{2 \left(a - b\right)}$

$x = \frac{- b + c \pm \sqrt{4 {a}^{2} + {b}^{2} + {c}^{2} - 4 a b + 2 b c - 4 a c}}{2 \left(a - b\right)}$

$x = \frac{- b + c \pm \sqrt{\left({b}^{2} + 2 b c + {c}^{2}\right) + \left(4 {a}^{2} - 4 a b - 4 a c\right)}}{2 \left(a - b\right)}$

$x = \frac{- b + c \pm \sqrt{{\left(b + c\right)}^{2} + 4 a \left(a - b - c\right)}}{2 \left(a - b\right)}$

Sep 10, 2016

The Roots are $1 , \mathmr{and} , \frac{c - a}{a - b}$.

#### Explanation:

Let $p \left(x\right) = \left(a - b\right) {x}^{2} + \left(b - c\right) x + \left(c - a\right) = 0$.

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

$p \left(x\right) = \left(a - b\right) + \left(b - c\right) + \left(c - a\right) = 0$.

$\therefore \left(x - 1\right) \text{ is a factor of }$p(x).

Hence, one root of $p \left(x\right) = 0 \text{ is 1.}$

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

The Product of roots of $A {x}^{2} + B x + C = 0$ ic $\frac{C}{A}$.

$\text{:. in our case, the products of roots=} \frac{c - a}{a - b}$, i.e.,

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

Thus, the soln. is, $x = 1 , x = \frac{c - a}{a - b}$.

Sep 10, 2016

$x = 1 , x = \frac{c - a}{a - b}$.

#### Explanation:

Let $p \left(x\right) = l {x}^{2} + m x + n = 0$, where,

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

Note that, $l + m + n = \left(a - b\right) + \left(b - c\right) + \left(c - a\right) = 0$, so,

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

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

$= \underline{l {x}^{2} - l x} - \underline{n x + n}$

$= l x \left(x - 1\right) - n \left(x - 1\right)$

$= \left(x - 1\right) \left(l x - n\right) = 0.$

Hnce, the roots are, $x = 1 , x = \frac{n}{l} = \frac{c - a}{a - b}$.

Enjoy Maths.!