Question

PLS ans it fast........​

Answer 1

First let's consider the standard quadratic equation $\sf{Ax^2+Bx+C=0.}$

We know the solution to this equation is given by,

$\longrightarrow\sf{x=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}}$

where $\sf{B^2-4AC}$ is known as the determinant which determines the nature of the roots of the equation.

$\longrightarrow\sf{D=B^2-4AC}$

Since the square root of discriminant is included for the solution as in the equation,

• to get real roots, the value of discriminant should be non - negative.

• to get rational roots, the discriminant should be in such a way that it can be expressed as a square of a real number.

Here we're given the equation,

$\longrightarrow\sf{(a-b)x^2+(b+c-a)x-c=0,\quad\quad a,\ b,\ c\in\mathbb{R},\quad\quad a\neq b}$

The discriminant is,

$\longrightarrow\sf{D=(b+c-a)^2-4(a-b)(-c)}$

$\longrightarrow\sf{D=(b+c-a)^2+4c(a-b)}$

$\longrightarrow\sf{D=b^2+c^2+a^2+2bc-2ab-2ac+4ac-4bc}$

$\longrightarrow\sf{D=a^2+c^2+b^2+2ac-2ab-2bc}$

$\longrightarrow\sf{D=(a+c-b)^2}$

Here we see that the discriminant is expressed as a square of a real number, since a, b, c are real numbers. Also it is non - negative.

$\longrightarrow\sf{D=(a+c-b)^2\geq0}$

Thus we can say that the roots of the equation $\sf{(a-b)x^2+(b+c-a)x-c=0}$ are rational, and the roots are,

$\longrightarrow\sf{x=\dfrac{a-b-c\pm(a+c-b)}{2(a-b)}}$

$\longrightarrow\sf{x=1\quad\quad OR\quad\quad x=\dfrac{c}{b-a},\quad a\neq b}$