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11 Mar 2019, 06:33
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
41% (01:18) correct
59%
(01:25)
wrong
based on 280
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GMATH practice exercise (Quant Class 12)
If \(x>0\), what is the sum of the roots of the equation \(\,{x^{\sqrt x }} = {x^2}\) ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
If \(x>0\), what is the sum of the roots of the equation \(\,{x^{\sqrt x }} = {x^2}\) ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
11 Mar 2019, 07:29
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IMO E ,
after compare , we get √x= 2
means x = 4 (as x>0)
and also by trial , we see 1 satisfy the equation.
so sum = 5
Posted from my mobile device
after compare , we get √x= 2
means x = 4 (as x>0)
and also by trial , we see 1 satisfy the equation.
so sum = 5
Posted from my mobile device
11 Mar 2019, 15:13
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fskilnik
\(x > 0\,\,,\,\,\,\,{x^{\sqrt x }} = {x^2}\,\,\,\left( * \right)\)
\(?\,\,:\,\,{\rm{sum}}\,\,{\rm{of}}\,\,{\rm{roots}}\)
\(x = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{inspection}}} \,\,\,\,\left\{ \matrix{\\
\,\,{x^{\sqrt x }} = {1^{\sqrt 1 }} = 1 \hfill \cr \\
\,\,{x^2} = {1^2} = 1 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{first}}\,\,{\rm{root}}\)
\(0 < x < 1\,\,\,{\rm{or}}\,\,\,x > 1\,\,\,:\,\,\,\,\,\)
\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,{x^{\sqrt x \, - \,2}} = 1 = {x^0}\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{base}}\,\, \ne \,\,0,1, - 1} \,\,\,\,\,\sqrt x - 2 = 0\,\,\,\, \Rightarrow \,\,\,\,\,x = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{second}}\,\,{\rm{root}}\)
\(? = 1 + 4\)
The correct answer is (E).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
