Question

If a,b,c ∈ R, show that the roots of the equation (a-b)x^2+ (b+c-a)x -c=0 are rational.

This question is from quadratic equations class 10 icse R.S AGARWAL exercise 5C questions 24 please friends i need urgent answers thank you so much

Answer 1

SOLUTION

TO PROVE

If a,b,c ∈ R, show that the roots of the below equation are rational

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

EVALUATION

Here the given equation is

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

Now Discriminant of the equation = D

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

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

$\sf{ = {(y - c)}^{2} + 4 yc \: \: \: where \: \: y = a - b }$

$\sf{ = {(y + c)}^{2} }$

$\sf{ = {(a - b + c)}^{2} } \geqslant 0$

Since a,b,c ∈ R

So the Discriminant is a perfect square number

Therefore the roots of the quadratic equation are rational

Hence proved

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Answer 2

$\underline{\textbf{Given:}}$

$\textsf{Equation is}$

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

$\underline{\textbf{To prove:}}$

$\textsf{Roots of the given equation are rational}$

$\underline{\textbf{Solution:}}$

$\textsf{Sum of the coefficients}}$

$\mathsf{=(a-b)+(b+c-a)+(-c)}$

$\mathsf{=0}$

$\therefore\mathsf{x=1\;is\;a\;root}$

$\textsf{By Synthetic dvision,}$

$\mathsf{\begin{array}{c|ccc}1&a-b&b+c-a&-c\\&&&\\&&a-b&c\\\cline{2-4}&a-b&c&\boxed{0}\end{array}}$

$\mathsf{Now,}$

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

$\implies\mathsf{x=\dfrac{-c}{a-b}}$

$\implies\textsf{The\;roots\;are\;1\;\;and\;\;\dfrac{-c}{a-b}}$

$\textsf{Hence, the roots of the equation are rational}$

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