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Problem: Find number of solutions of $|2x^2-5x+3|+x-1=0$

Solution: Case 1: When $2x^2-5x+3 \geq 0$

Then we get, $2x^2-5x+3+x-1=0$

x=1,1

Case 2: When $2x^2-5x+3 < 0$

Then we get, $-2x^2+5x-3+x-1=0$

x=1,2

In both cases, common value of x is 1

Hence solution is x=1

Am I doing right ??

Somebody told me solution is x=2

Community
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rst
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  • The simplest approach is to test all solutions in the original problem. Either or both may have produced extraneous solutions because although the equation has a solution, that value of $x$ does not place it in that case. – vadim123 Jun 24 '13 at 14:10
  • The absolute value is equal to $1-x$: you can accept only values such that $1-x\ge 0 ,$, i.e. $, x \le 1$. – Tony Piccolo Jun 24 '13 at 14:16

2 Answers2

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Nope, if you have $|f(x)| + g(x) = 0$ $(1)$ you solve it as follows:

  1. assume that $f(x)\geq 0$, find solutions of $f+g = 0$ - say $x_1$. Check whether $f(x_1)\geq 0$. If it holds, then $x_1$ solves $(1)$ as well. If it does not hold, it's not a solution of the original equation.

  2. assume $f(x)<0$, find solutions of $g-f = 0$ and check whether for them $f<0$.

In your case: for $f+g = 0$ you get a solution $x = 1$, which satisfies $f(1)\geq 0$, so that this is indeed one of the solutions of the original equation. For the case $f<0$ you get $x = 1$ or $x = 2$. We don't have to check $x = 1$ since we know it's already a solution, so we only check whether $f(2)<0$. Since the latter does not hold: $2\cdot 2^2 -5\cdot 2+3 = 1>0$, we obtain that $x=2$ is not a solution, and hence $1$ is the only solution.

SBF
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  • Thanks,you really cleared my doubts – rst Jun 24 '13 at 14:19
  • According to case $(2)$, $x=1 $ is not solution of $f $ i.e $f(1)=0$ (which is not less than $0$) – rst Jun 24 '13 at 14:48
  • Well, that doesn't matter anymore @rst: to be a solution if the original equation, for $x$ it is necessary and sufficient to be a solution of at least one of versions, and to satisfy the condition of the corresponding version. If it happens to be a solution in both versions, but satisfy conditions only of the first one, it is still a solution. It doesn't have to satisfy both constraints, even if it happens to be a solution of both versions. – SBF Jun 24 '13 at 14:51
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$$|2x^2-5x+3|+x-1= \begin{cases} 2x^2-5x+3+x-1 & x\ge\frac32 \quad \text{and} \quad x\le1 \\ -(2x^2-5x+3)+x-1 & 1\lt x\lt\frac32\\ \end{cases}\Rightarrow\begin{cases} 2x^2-4x+2 & x\ge\frac32 \quad \text{and} \quad x\le1 \\ -2x^2+6x-4 & 1\lt x\lt\frac32\\ \end{cases}$$

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M.H
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