Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</= 0,

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isaiah45798
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Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</= 0,

Postby isaiah45798 » Mon Jul 18, 2011 3:04 am

Can someone please help or guide me with these problems.
Thanks in advance.
less than or equal to = </=

(x+3)(x-3)>0

x^2-2x-15</= 0

sin2x=sinx, 0</=x</=2pi

logx+log(x-3)=1

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maggiemagnet
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Re: Help with solving equations

Postby maggiemagnet » Tue Jul 19, 2011 4:55 pm

isaiah45798 wrote:(x+3)(x-3)>0

x^2-2x-15</= 0

To learn how to solve polynomial inequalities, try here.

The first one is already factored, so it's halfway done. For the second one, the first step would be to factor the quadratic.

isaiah45798 wrote:sin2x=sinx, 0</=x</=2pi

To learn how to solve trig equations, try here.

A good first step would be to move the sine to the left-hand side, apply the double-angle identity to the first sine, and then factor. Then sovle each factor.

isaiah45798 wrote:logx+log(x-3)=1

To learn how to solve log equations, try here.

A good first step would be to use log rules to combine the terms on the left-hand side. Then covert to the exponential form, and solve the rational equation you get out.

If you get stuck, please write back showing how far you've gotten. Thanks!

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isaiah45798
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Re: Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</=

Postby isaiah45798 » Tue Jul 19, 2011 9:59 pm

Thank you it all helped. I am stuck however on the trig equation. I am not sure what to do after I get 2sin(x)cos(x)=sin(x).

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maggiemagnet
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Re: Help with solving equations: (x+3)(x-3)>0, x^2-2x-15</=

Postby maggiemagnet » Wed Jul 20, 2011 12:02 am

isaiah45798 wrote:Thank you it all helped. I am stuck however on the trig equation. I am not sure what to do after I get 2sin(x)cos(x)=sin(x).

A good first step would be to move the sine to the left-hand side, apply the double-angle identity to the first sine, and then factor. Then solve each factor. :wink:

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