Question Number 222299 by fantastic last updated on 22/Jun/25
$${For}\:{what}\:{value}\:{of}\:\:{k}\:\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${will}\:{have}\:{same}\:{value}\:{but}\:\:{with}\:{opposite}\:{symbol}\left({like}\:{x}={a}\:{and}\:−{a}\right) \\ $$$${i}\:{mean}\:{the}\:{two}\:{valuea}\:{of}\:{x}\:{will}\:{be}\:{this}\:{type} \\ $$$${x}=\mathrm{2}\:{and}\:−\mathrm{2}\left({both}\:\mathrm{2}\:{but}\:{opposite}\:{symbols}\right) \\ $$
Answered by mr W last updated on 22/Jun/25
$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{3}{k}−\mathrm{1}}{{k}+\mathrm{1}}\right){x}+\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}=\mathrm{0} \\ $$$$\Rightarrow{sum}\:{of}\:{roots}=\mathrm{2}\left(\frac{\mathrm{3}{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)=\mathrm{0}\: \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{3}}\:\checkmark \\ $$
Commented by fantastic last updated on 22/Jun/25
$${sir}\:{how}\:{did}\:{you}\:{get}\:{the}\:\mathrm{2}^{{nd}} \:{line}? \\ $$$${please}\:{tell}\:{me} \\ $$
Commented by fantastic last updated on 22/Jun/25
$${thanks}\:{sir} \\ $$
Commented by mr W last updated on 22/Jun/25
![x^2 −2x=4x(((k−1)/(k+1)))−((k−1)/(k+1)) x^2 −2[1+2(((k−1)/(k+1)))]x+((k−1)/(k+1))=0 x^2 −2(((3k−1)/(k+1)))x+((k−1)/(k+1))=0](https://www.tinkutara.com/question/Q222313.png)
$${x}^{\mathrm{2}} −\mathrm{2}{x}=\mathrm{4}{x}\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)−\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\left[\mathrm{1}+\mathrm{2}\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}\right)\right]{x}+\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{3}{k}−\mathrm{1}}{{k}+\mathrm{1}}\right){x}+\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}=\mathrm{0} \\ $$