Help with solving equations for x

[isaiah45798]

Homework Statement

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

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

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

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

4 logx+log(x-3)=1

The Attempt at a Solution

OK so for number 1 I do not know where to start.
For number two I factor it and I'm in the same spot as number 1.
As far as number 3 is concerned I do not know where to start.
Number 4 I know that I have to condense the logarithms into one but I'm not sure what to do afterwards.

 

[vela]

For #1, multiply the lefthand side out.

For #2, complete the square.

Note for #1 and #2, you could also look at the sign of each factor. If ab>0, then you must have a>0 and b>0 OR a<0 and b<0. What do you need for ab<0?

For #3, use the trig identity sin 2x = 2 sin x cos x.

For #4, what's the inverse of the logarithm function?

 

[cupid.callin]

#1 try ways curve method ...

 

[isaiah45798]

thank you both for the help

 

[vela]

Did you figure them out?

 

[zaldar]

On number 1 can you just not set both factors equal to greater than 0 and solve for x?

Ah completing the square! so much fun.

 

[SammyS]

On number 1 can you just not set both factors equal to greater than 0 and solve for x?

No.

For some values of x, (x-3) and (x+3) have opposite sign. For these values, is (x-3)(x+3) positive, or is it negative?

For other values of x, (x-3) and (x+3) have like signs. For these values, is (x-3)(x+3) positive, or is it negative?

 

[cupid.callin]

On number 1 can you just not set both factors equal to greater than 0 and solve for x?

For these type of questions you can only ignore those terms to find range which are always >0 no matter what is x

 

[zaldar]

hmm so on 1 after multiplying it out you then complete the square to solve for x since that will give you an actual number on the right side?

Interesting wrinkle since as long as it is a number this is how you solve these type of problems factor and then set equal to the number or 0 and solve for x.

Though wait it has to be greater than 0 so does that not mean that both x's have to be positive? So x can be greater than 3 (from x -3) or greater than -3 from x +3

if x is -2 then x+3 would be 1, x-3 would then be ug -5 ok so that doesn't work..interesting.

 

[cupid.callin]

#1
completing square will give you ... x² - 9 = 0 or x² = 9

now just take square root

but don't forget [itex]\sqrt{x^2} \ = \ |x| [/itex] and not just x