18 Consider the equation $$ x^{2}-2 x+r=0 $$, where $$ r $$ is a real parameter. $$ 1^{\circ} $$ Calculate $$ r $$ so that the roots $$ x_{1} $$ and $$ x_{2} $$ exist. $$ 2^{\circ} $$ Calculate $$ r $$ so that : $$ \begin{array}{lll} ext { a) } \frac{1}{x_{1}}+\frac{1}{x_{2}}=-\frac{1}{2} & ext {; b) } x_{1}^{2}+x_{2}^{2}=6 & ext {; ce } \frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=1 \ ext { d) }\left(x_{1}-x_{2} ight)^{2}=4 & ext {; e) } \frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=3\end{array} $$

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18 Consider the equation $$ x^{2}-2 x+r=0 $$, where $$ r $$ is a real parameter. $$ 1^{\circ} $$ Calculate $$ r $$ so that the roots $$ x_{1} $$ and $$ x_{2} $$ exist. $$ 2^{\circ} $$ Calculate $$ r $$ so that : $$ \begin{array}{lll}	ext { a) } \frac{1}{x_{1}}+\frac{1}{x_{2}}=-\frac{1}{2} & 	ext {; b) } x_{1}^{2}+x_{2}^{2}=6 & 	ext {; ce } \frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=1 \ 	ext { d) }\left(x_{1}-x_{2}ight)^{2}=4 & 	ext {; e) } \frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=3\end{array} $$

Nunez Washington · Accepted Answer

For the roots to exist, $$ r \leq 1 $$. For the specific conditions: a) $$ r = -4 $$, b) $$ r = -1 $$, c) $$ r = -1 + \sqrt{5} $$ or $$ r = -1 - \sqrt{5} $$, d) $$ r = 0 $$, e) $$ r = -\frac{2}{3} $$.

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