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Do you just change the absolute value signs with parentheses?

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3 Answers 3

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Let $|x + 1| = A$. Then we have the quadratic equation $2A^2 = 3A + 2$, or $2A^2 - 3A - 2 = 0$. Factoring, we have that $(2A + 1)(A - 2) = 0$. This yields $A =-1/2$ and $A = 2$. $A = -1 / 2$ is not possible, so$ A = |x + 1| = 2$, solving, we get $x + 1 = -2$ or $2$, and $x = -3$ or $1$.

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    $\begingroup$ Please use MathJax for formatting. Your answers will be better received. $\endgroup$
    – Shailesh
    Commented Jan 7, 2016 at 2:21
  • $\begingroup$ Thank you for formatting my solution. $\endgroup$
    – K. Jiang
    Commented Jan 7, 2016 at 2:24
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Hint: What if, instead of $\lvert x+1\rvert$, the equation had $y$? $$ 2y^2=3y+2 $$

How would you solve it then?

Once you've found values for $y$ that satisfy this equation, "remember" that $y=\lvert x+1\rvert$.

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  • $\begingroup$ Thank you! Should've noticed that. I was too focused on the absolute value being squared. $\endgroup$
    – user303287
    Commented Jan 7, 2016 at 2:55
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Hint:

Set $t=\lvert x+1\rvert$, solve the quadratic equation for $t$, then solve $\lvert x+1\rvert=t\;$ (this last equation implies $t$ is a non-negative solution of the quadratic equation).

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