Question

24. If a, b, c belongs to R, show that the roots of the equation (a - b)x2 + (b + c - a)x - c = 0 are
rational.
Hint : D = (a + c - 6)2 > 0​

Answer 1

Quadratic equation

Solution.

The given equation is

$\quad (a-b)x^{2}+(b+c-a)x-c=0$

Now, discriminant $(D)$

$\quad=(b+c-a)^{2}-4(a-b)(-c)$

$\quad=b^{2}+c^{2}+a^{2}+2bc-2ca-2ab+4ca-4bc$

$\quad=a^{2}+c^{2}+b^{2}+2ca-2ab-2bc$

$\quad=(a+c-b)^{2}$

$\because a,\:b,\:c\in\mathbb{R}$, $(a+c-b)^{2}\in\mathbb{R}$ and $>0$.

$\because D$ is a perfect square, the roots of the given equation are rational.

Thus proved.

Finding the roots of the given equation.

The given equation is

$\quad (a-b)x^{2}+(b+c-a)x-c=0$

$\Rightarrow (a-b)x^{2}-(a-b-c)x-c=0$

$\Rightarrow (a-b)x^{2}-(a-b)x+cx-c=0$

$\Rightarrow (a-b)x(x-1)+c(x-1)=0$

$\Rightarrow (x-1)\{(a-b)x+c\}=0$

Either $x-1=0$ or $(a-b)x+c=0$

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

Thus the roots are

$\quad \color{red}{x=1,\:-\frac{c}{a-b}}$.

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