Question 213355
Can you please help me with this equation?


{{{2m^(2/3)+3m^(1/3)-2=0}}} 


Using the factoring approach we have the following


Step 1.    Let {{{ x= m^(1/3)}}}

Then the {{{2m^(2/3)+3m^(1/3)-2=0}}}  becomes  an quadratic equation in x  since {{{m^(2/3)}}}={{{(m^(1/3))^2}}}


{{{2x^2+3x-2=0}}} 


Step 2.  After trial and trial,  we have


{{{0=2x^2+3x-2=(2x-1)(x+2)}}}  we can verify using the FOIL method


Step 3.  Now we have 2x-1=0 and x+2=0


{{{2x-1=0}}}


Add 1 to both sides to isolate x terms on left side


{{{2x-1+1=0+1}}}


{{{2x=1}}}


Divide 2 on both sides to get x


{{{2x/2=1/2}}}


{{{x=1/2}}}


Step 4.  Now let's work 


{{{x+2=0}}}


Subtract 2 from both sides to get x


{{{x+2-2=0-2}}}

{{{x=-2}}}


Step 5. Based on the above steps, x=0.5 and x=-2


Step 6.  Since {{{x= m^(1/3)}}} then


{{{0.5=m^(1/3)}}}


{{{0.5^3=(m^(1/3))^3}}}


{{{0.125=m}}}


Now for {{{x=-2}}}


{{{-2=m^(1/3)}}}


{{{(-2)^3=(m^(1/3))^3}}}


{{{m=-8}}}


Step 5.  ANSWER:  So m = 0.125 and -8.


The following steps show how to solve the above quadratic equation using the quadratic formula.


Step 1.   Let {{{ x= m^(1/3)}}}


Then the {{{2m^(2/3)+3m^(1/3)-2=0}}}  becomes  an quadratic equation in x  since {{{m^(2/3)}}}={{{(m^(1/3))^2}}}


{{{2x^2+3x-2=0}}} 


Step 2.  Then we use the quadratic formula to {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=2, b=3 and c=-2.


*[invoke quadratic "x", 2, 3, -2 ]


Step 3 Based on the above steps, x=0.5 and x=-2


Step 4.  Since {{{x= m^(1/3)}}} then


{{{0.5=m^(1/3)}}}


{{{0.5^3=(m^(1/3))^3}}}


{{{0.125=m}}}


Now for {{{x=-2}}}


{{{-2=m^(1/3)}}}


{{{(-2)^3=(m^(1/3))^3}}}


{{{m=-8}}}


Step 5.  ANSWER:  So m = 0.125 and -8.


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J