. Consider the substitution u = (x+1). Then the equation turns out to be 
then its solutions are given by the expression ![x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}](/tpl/images/0690/6009/161c4.png)
![u_1 = \frac{4+\sqrt[]{(-4)^2-4(2)}}{2} = 2 +\sqrt[]{2}](/tpl/images/0690/6009/fe380.png)
![u_2 = \frac{4-\sqrt[]{(-4)^2-4(2)}}{2} = 2 -\sqrt[]{2}](/tpl/images/0690/6009/eac1e.png)
![x_1 = 2 +\sqrt[]{2}-1 = 1 + \sqrt[]{2}](/tpl/images/0690/6009/73280.png)
![x_2 =2 -\sqrt[]{2} -1 = 1 - \sqrt[]{2}](/tpl/images/0690/6009/6d899.png)
ั
4(x +1)
Mathematics, 20.06.2020 03:57 jtmoney10
Solve (x + 1)2 - 4(x + 1) + 2 = 0 using substitution.
U=
ั
4(x +1)
O x +1
(x + 1)2
Answers: 3
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You know the right answer?
Solve (x + 1)2 - 4(x + 1) + 2 = 0 using substitution.
U=
ั
4(x +1)
ั
4(x +1)
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