Darkplayer
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- Sep 6, 2008
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I just cant solve this equation, i tried and tried but i just can´t! Can somone please help me?
(x+1,5)(x+1,5)+((3/x)+2)((3/x)+2)=25
(x+1,5)(x+1,5)+((3/x)+2)((3/x)+2)=25
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Havin some trouble solving (x + 1,5)^2 + ( (3/x) + 2)^2 = 25
- Thread starter Darkplayer
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D
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Re: Havin some trouble with equation!
I assume '," (comma) stands for decimal point. Is that correct?
Does your problem look like:
\(\displaystyle (x \, + \, 1.5)\cdot (x \, + \, 1.5) \, + \, (\frac{3}{x} \, + \, 2) \cdot (\frac{3}{x} \, + \, 2)) \, = \, 25\)
If it does, then
\(\displaystyle (x \, + \, \frac {3}{2})\cdot (x \, + \, \frac {3}{2}) \, + \, (\frac{3 \, + 2x}{x}) \cdot (\frac{3 \, + 2x}{x}) \, = \, 25\)
\(\displaystyle \frac{(3 \, + 2x)^2}{4} \, + \, \frac{(3 \, + 2x)^2}{x^2} = \, 25\)
\(\displaystyle (3 \, + 2x)^2 \cdot x^2 \, + \, (3 \, + 2x)^2 \cdot 4 = \, 25\cdot 4 \cdot x^2\)
can you continue now.....
Darkplayer said:I just cant solve this equation, i tried and tried but i just can´t! Can somone please help me?
(x+1,5)(x+1,5)+((3/x)+2)((3/x)+2)=25
I assume '," (comma) stands for decimal point. Is that correct?
Does your problem look like:
\(\displaystyle (x \, + \, 1.5)\cdot (x \, + \, 1.5) \, + \, (\frac{3}{x} \, + \, 2) \cdot (\frac{3}{x} \, + \, 2)) \, = \, 25\)
If it does, then
\(\displaystyle (x \, + \, \frac {3}{2})\cdot (x \, + \, \frac {3}{2}) \, + \, (\frac{3 \, + 2x}{x}) \cdot (\frac{3 \, + 2x}{x}) \, = \, 25\)
\(\displaystyle \frac{(3 \, + 2x)^2}{4} \, + \, \frac{(3 \, + 2x)^2}{x^2} = \, 25\)
\(\displaystyle (3 \, + 2x)^2 \cdot x^2 \, + \, (3 \, + 2x)^2 \cdot 4 = \, 25\cdot 4 \cdot x^2\)
can you continue now.....
DR._Glockman
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- Jun 10, 2008
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Do you have a TI-89? If so Then F2-1(solve)((x+1.5)^2 + (3/x+2)^2=25,x)
Answers: x=2.1987..., x=1.5, x=-.4357..., x=-6.2630... (all reals)
Note: Following Subhotosh Khan instructions (after a little algebraic manipulation), you''ll get 4x^4+12x^3-75x^2+48x+36 =0
If you factor this you'll get the above answers. x = 1.5 is the only rational number, the other three I am assuming are irrationals.
Answers: x=2.1987..., x=1.5, x=-.4357..., x=-6.2630... (all reals)
Note: Following Subhotosh Khan instructions (after a little algebraic manipulation), you''ll get 4x^4+12x^3-75x^2+48x+36 =0
If you factor this you'll get the above answers. x = 1.5 is the only rational number, the other three I am assuming are irrationals.