Linear Equations with a Single Unknown: Variations
Substitution
Consider the equation \[3x^4+5=20 \tiny.\] This equation is not linear in unknown #x#. But if we rewrite the equation with the new unknown #y=x^4#, the linear equation #3y+5=20# emerges with unknown #y#. This one can be solved: #y=5#. We then still have to solve the equation #x^4=5#. The theory of Higher roots gives #x=\sqrt[4]{5}\lor x=-\sqrt[4]{5}#. With which the original equation is solved.
The solution method for an equation with one unknown by replacing the unknown by another one which can be expressed in terms of the original unknown, is called substitution.
Another way to describe this process is a transition to two equation with two unknowns. In the example #3x^4+5=20# the new system is #3y+5 = 20\land y = x^4#.
The trick is to find the substitution that makes sure
- the new equation with unknown #y# can be solved with less difficulty than the original equation, and
- the solutions of the second equation with unknown #x# can be discoverde with given #y#.
Give your answer in the form #x=a# or #x=a \lor x=b# for suitable #a# and #b#.
We write #y=(x-5)^{4}#, such that the equation #2 (x-5)^{4}=6# can be written as # 2 y=6 # with unknown #y#. The solution to this equation is #y=3#.
Now we must solve the equation #(x-5)^{4} = 3# with unknown #x#. The theory of Higher roots gives # {x-5=\sqrt[4]{3} \lor x -5=-\sqrt[4]{3}} #. Hence the answer is #x=\sqrt[4]{3}+5 \lor x =-\sqrt[4]{3}+5#.
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