Need Some Math Help

  • Thread starter kahi
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  • #1
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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
 

Answers and Replies

  • #2
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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
dextercioby
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kahi said:
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

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

Daniel.

PS.And the first eq.factors as well:(x-3)(x-1)=0.
 
  • #4
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Factoring hurts my brain...
 
  • #5
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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
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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
dextercioby
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kahi said:
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

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

Daniel.
 
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