SOLUTION: Hi, can you help me solve an Equation Quadratic in Form?
the Equation is {{{6x^-1-5x^(-1/2)+1=0}}}
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Question 69705This question is from textbook Colelge Algebra Essentials
: Hi, can you help me solve an Equation Quadratic in Form?
the Equation is
This question is from textbook Colelge Algebra Essentials
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
solve the Equation 6x^-1-5x^(-1/2)+1=0
(3x^(-1/2) - 1)(2x^(-1/2)-1)=0
3x^(-1/2) = 1 or 2x^(-1/2)=1
x^(-1/2)=1/3 or x^(-1/2)= 1/2
Square both sides to get:
x^-1=1/9 or x^-1= 1/4
x=9 or x=4
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Check the answers in the original equation:
6x^-1-5x^(-1/2)+1=0
If x=9,
6(1/9)-5(1/3)+1=0
2/3-5/3+3/3=0
0=0
--------------
If x=4,
6(1/4)-5(1/2)+1=0
3/2-5/2+2/2=0
0=0
-------------
Both answers check out so x=9 or x=4
Cheers,
Stan H.
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