simple quadratic

All of the quantities are within absolute value signs. Thus if any of those quantities within the absolute value signs are anything other than 0, then the equation will not be true because anything in absolute value signs are either 0 or positive. Therefore x - 1 must equal zero.

Thus we can set $x - 1 = 0$ , so x must be 1.

Note that if you set ${ x }^{ 2 }-x-2=0$ equal to zero, the -2 that we get as a solution is extraneous.

$X = 1$

Since the term $(x-1)^2$ can be rewritten as the product of identical terms in absolute value signs, and $\left | x^2 +x -2 \right |= \left | x+2 \right |\left | x-1 \right |$ , then we can rewrite the equation to the following form.

$(x-1)^2 + \left | x-1 \right | + \left | x^2 +x -2 \right | = \left ( \left | x-1 \right | \right )^2 + \left | x-1 \right | + \left | x+2 \right |\left | x-1 \right |$

Factoring out $\left | x-1 \right |$ , we can solve for x.

$\left | x-1 \right |(\left | x-1 \right | + 1 + \left | x+2 \right |) = 0$

Since the sum of the terms in parentheses can't be equal to 0, then it must be that $x = 1$ .

Let a = (x-1)^2, b = | x - 1 |, c = | x^2 + x -2 |. All terms (a, b, c) are positive so their summation can only be zero if each is zero. x=1 yields zero for each of the terms a,b, and c.

Since the RHS is 0 the sum of the terms on the LHS is 0 and therefore all three terms must be 0. And x = 1 is the only value that solves the equation.

So easy ... Just substitue $x=1$ into the equation : $(x-1)^2+|x-1|+|(x^2)+x-2|=0$ $(1-1)^2+|1-1|+|1^2+1-2|=0$

$(0)^2+|0|+|1-1|=0$

$0=0$

So we get x = 1

Each and every part inside the modulus sign is >=0 so we can say that it will be equal to 0 if we have (x-1)²=0

As all the terms are positive and summing up of the positive terms can not be zero, so we have to make all terms equal to zero.....So we can say x-1=0 =>x=1.

Since each quantities are positive each of them must be 0. The only number for each quantity which makes them 0 is 1. x-1=0 and x=1