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From Förberedande kurs i matematik 1

If we write the equation as

ex2+ex=4 

we see that x appears only in the combination ex and it is therefore appropriate to treat ex as a new unknown in the equation and then, when we have obtained the value of ex, we can calculate the corresponding value of x by simply taking the logarithm.

For clarity, we set t=ex, so that the equation can be written as

t2+t=4

and we solve this second-degree equation by completing the square,

t2+t=t+212−212=t+212−41 

which gives

t+212−41=4t=−21217 

These two roots give us two possible values for ex,

ex=−21−217  or ex=−21+217 

In the first case, the right-hand side is negative and because " e raised to anything" can never be negative, there is no x that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because 171  ) and we can take the logarithm of both sides to obtain

x=ln217−21 

NOTE: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting t=217−21  into the equation t2+t=4,

LHS =

\displaystyle \begin{align} & =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\ & =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\ \end{align}

\displaystyle = RHS.