Need Some Math Help

1. Jan 1, 2005

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 completly forgotten and have left my book elsewhere

2. Jan 1, 2005

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.

3. Jan 1, 2005

dextercioby

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.

4. Jan 1, 2005

Yapper

Factoring hurts my brain...

5. Jan 1, 2005

kahi

ok thats 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

6. Jan 1, 2005

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.

7. Jan 2, 2005

dextercioby

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

Daniel.

Last edited: Jan 2, 2005