Question 1192197
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(a) Solve the equation 4y^2 + 3y - 1 = 0.
(b) Use your answer to part (a) to solve the equation 4(2^x)^2 + 3(2^x) - 1 = 0.
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(a)  By using the quadratic formula or by factoring, the roots of the first equation are  -1  or  1/4.



(b)  Considering  {{{2^x}}}  as a new variable y, we see that either  {{{2^x}}} = -1  or  {{{2^x}}} = 1/4.


     First equation has no real solutions.

     Second equation  {{{2^x}}} = 1/4  has the unique solution  x= -2.

     Therefore, the only solution for part (b) is x= -2.
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Solved.