The roots of |x - 2{|^2} + |x - 2| - 6 = 0are | vedclass

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Question

(a) When $x < 2,$ ${(x - 2)^2} - (x - 2) - 6 = 0$

$ \Rightarrow {x^2} - 4x + 4 - x + 2 - 6 = 0$$ \Rightarrow $ ${x^2} - 5x = 0$

$ \Rightarrow

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$0, 4$

Vedclass · Answer

(a) When $x < 2,$ ${(x - 2)^2} - (x - 2) - 6 = 0$

$ \Rightarrow {x^2} - 4x + 4 - x + 2 - 6 = 0$$ \Rightarrow $ ${x^2} - 5x = 0$

$ \Rightarrow $ $x(x - 5) = 0$ $ \Rightarrow $ $x = 0$

When $x \ge 2$; ${(x - 2)^2} + (x - 2)\, - 6 = 0$

$ \Rightarrow {x^2} - 4x + 4 + x - 2 - 6 = 0$

$ \Rightarrow $ ${x^2} - 3x - 4 = 0$

$ \Rightarrow $ ${x^2} - 4x + x - 4 = 0$

$ \Rightarrow $ $(x - 4)(x + 1) = 0$

$ \Rightarrow $ $x = 4$.

The roots of $|x - 2{|^2} + |x - 2| - 6 = 0$are

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