Find Solutions for Math Equations: Pretrig/Calc Review

[kahi]

I am doing a pretrig/calc review guide and it is asking me to find all the solutions for each equation.

-3x=x^2-4

that is the first equation

y^4-2y^2=-y^3

is another one

any help on what I am supposed to do...I have completely forgotten and have left my book elsewhere

 

[Yapper]

For the first problem add 3x to both sides so you end up with x^2+3x-4=0 with that use the quadratic formula [-b+/-(b^2-4ac)^(1/2)]/2a

For the second one I would try moving it all to one side and factoring if you can.

 

[dextercioby]

I am doing a pretrig/calc review guide and it is asking me to find all the solutions for each equation.
-3x=x^2-4
that is the first equation
y^4-2y^2=-y^3
is another one
any help on what I am supposed to do...I have completely forgotten and have left my book elsewhere

The second has an immediate factorization;
y^{4}+y^{3}-2y^{2}=0\Rightarrow y^{2}(y^{2}+y-2)=0
,which has a double solution y=0 and the other 2 are found by solving the quadratic
y^{2}+y-2=0

Daniel.

PS.And the first eq.factors as well:(x-3)(x-1)=0.

 

[Yapper]

Factoring hurts my brain...

 

[kahi]

ok that's what I thought ...but I am unsure with

2x^5=32x^3

and then the log stuff like

2logx=log25 I think this one is (2log5)?

or

lnx+ln(x+2)=4

 

[Yapper]

2x^3(16-x^2)=0
2x^3(4+x)(4-x)=0 difference of squares
x={0,0,0,4,-4}

I think.

 

[dextercioby]

ok that's what I thought ...but I am unsure with
2x^5=32x^3
and then the log stuff like
2logx=log25 I think this one is (2log5)?
or
lnx+ln(x+2)=4

2 \lg x =\lg25\Rightarrow 2\lg x=2\lg5\Rightarrow x=5
\ln x+ \ln(x+2) =4\Rightarrow \ln[x(x+2)]=4\Rightarrow x(x+2)=e^{4}
,which is a second order algebraic equation which can be solved.

Daniel.