For what values of c does the following equation have at least one real solution? $\displaystyle cx^2+sqrt(c) x = c$
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Originally Posted by Saphira For what values of c does the following equation have at least one real solution? $\displaystyle cx^2+\sqrt{c}x=c$ You want $\displaystyle (\sqrt{c})^2-4(c)(-c)\ge0$
Are we assuming $\displaystyle c \ne 0$? Are we assuming $\displaystyle c \in \mathbb{R}$?
The value of c is not specified, so I'm guessing, yes.
Originally Posted by Plato You want $\displaystyle (\sqrt{c})^2-4(c)(-c)\ge0$ Thanks!
Okay, have you considered the Quadratic Formula and the implications of the radical?
Originally Posted by Saphira For what values of c does the following equation have at least one real solution? $\displaystyle cx^2+sqrt(c) x = c$ 1. c is positive since otherwise $\displaystyle \sqrt{c}$ is not real and I presume we want it to be real. 2. Rewrite the equation: $\displaystyle cx^2+\sqrt{c}x-c=0$ What does Descartes rule of signs tell you when $\displaystyle c>0$. CB
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