How do you solve ${(c + 1)}^{2} - 4 (c + 1) + 3 = 0$ $(c + 1) ^ { 2} - 4( c + 1) + 3= 0$ ?

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Answer 1 · 2018-06-26T11:30:53

We can view this as a quadratic equation in $c + 1$ $c+1$ and factor that way:

$0 = {(c + 1)}^{2} - 4 (c + 1) + 3 = ((c + 1) - 3) ((c + 1) - 1) = (c - 2) c$ $0=(c+1)^2-4(c+1)+3=( (c+1)-3)((c+1)-1)=(c-2)c$

$c = 2 or c = 0$ $c=2 or c=0$

Answer 2 · 2018-06-26T11:32:39

$c = {0, 2}$ $c = {0,2}$

Let $c + 1 = x$ $c+1 = x$

$\Rightarrow x^{2} - 4 x + 3 = 0$ $=> x^2 -4x +3 = 0$

$\Rightarrow (x - 3) (x - 1) = 0$ $=> (x-3)(x-1) = 0$

$\Rightarrow x = 3$ $=> x = 3$

$\Rightarrow x = 1$ $=> x =1$

Hence

$c + 1 = 3 \Rightarrow c = 2$ $c + 1 = 3 => c = 2$

$c + 1 = 1 \Rightarrow c = 0$ $c+1 = 1 => c = 0$

$c = {2, 0}$ $c = { 2,0}$

How do you solve (c + 1) ^ { 2} - 4( c + 1) + 3= 0(c+1)2−4(c+1)+3=0(c + 1) ^ { 2} - 4( c + 1) + 3= 0?