$$e^x=m(m+1),\; m<0$$ I draw the graph of R.H.S and LHS and got it intersect at one point . But i dont know whether my procedure is right or wrong. Plz help me. The options are 1) no real root. 2) exactly one. 3) two real roots.
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Welcome to Math SE! Please format your question using MathJax. You can also show some effort by showing your work and clarifying where you are stuck on. – Toby Mak Oct 28 '17 at 05:21
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The most straightforward way would be to look at the graph of $e^x - m(m+1)$. Since $e^x$ is strictly increasing, what does this tell you about the roots? – Toby Mak Oct 28 '17 at 05:23
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Thank you Toby Mak. Since x and m are different variables how would be the graph drawn for e^x -m(m+1). This is the matter of trouble for me. – Shagun Oct 28 '17 at 05:51
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Plot the curves for some values of $m<0$, e.g. $m=-2$ or $m=-1/2$. What do you find?
Since $e^x$ is strictly increasing, there is at most one real solution. If there is a real solution then it is $x=\ln(m(m+1))$. Now you have to look for values of $m$ with $m(m+1) > 0$. I guess you can continue.
gammatester
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