+26400 1. |x2-1|=|x-1|
\(\begin{array}{rcll} |x^2-1| &=& |x-1| \qquad & (x^2-1) = (x-1)(x+1) \\ (|(x-1)(x+1)|)^2 &=& (|x-1|)^2 \qquad & (\text{square both sides}) \\ (~(x-1)(x+1)~)^2 &=& (x-1)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ (x-1)^2(x+1)^2 &=& (x-1)^2 \\ (x-1)^2(x+1)^2-(x-1)^2 &=& 0\\ (x-1)^2\left[(x+1)^2-1 \right] &=& 0\\ (x-1)^2(x^2+2x+1-1) &=& 0\\ (x-1)^2(x^2+2x) &=& 0\\ (x-1)^2\cdot x \cdot (x+2) &=& 0\\ (x-1)\cdot (x-1)\cdot x \cdot (x+2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & x = 0 \\ 3. & x+2 = 0 && x = -2 \end{array}\)
2. 3|x-2|=2|x-3|
\(\begin{array}{rcll} 3|x-2| &=& 2|x-3| \\ (3|x-2|)^2 &=& (2|x-3|)^2 \qquad & (\text{square both sides}) \\ 3^2(x-2)^2 &=& 2^2(x-3)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ 9(x-2)^2 &=& 4(x-3)^2 \\ 9(x^2-4x+4) &=& 4(x^2-6x+9) \\ 9x^2-36x+36 &=& 4x^2-24x+36 \\ 9x^2-36x &=& 4x^2-24x\\ 5x^2-12x &=& 0\\ x\cdot(5x-12) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 0 \\ 2. & 5x-12=0 && x=\frac{12}{5} \end{array}\)
3. |4-x|=5
\(\begin{array}{rcll} |4-x| &=& 5 \\ (|4-x|)^2 &=& (5)^2 \qquad & (\text{square both sides}) \\ (4-x)^2 &=& 25 \qquad & (\pm\sqrt{}) \\ 4-x &=& \pm 5\\ x &=& 4 \pm 5 \\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 4+5 && x = 9 \\ 2. & x = 4-5 && x = -1 \end{array}\)
4. |2x-7|=|x-5|
\(\begin{array}{rcll} |2x-7| &=& |x-5| \\ (|2x-7|)^2 &=& (|x-5|)^2 \qquad & (\text{square both sides}) \\ (2x-7)^2 &=& (x-5)^2 \\ 4x^2-28x+49 &=& x^2-10x+25\\ 3x^2-18x+24 &=& 0\qquad & :3\\ x^2-6x+8 &=& 0\\ (x-4)(x-2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-4=0 && x = 4 \\ 2. & x-2=0 && x=2 \end{array}\)
5. |x2-x|=|x2-1|
\(\begin{array}{rcll} |x^2-x| &=& |x^2-1| \\\\ \qquad x^2-x = x(x-1) \\ \qquad (x^2-1) = (x-1)(x+1) \\\\ |x(x-1)| &=& |(x-1)(x+1)| \\ (~|x(x-1)|~)^2 &=& (~|(x-1)(x+1)|~)^2 & (\text{square both sides}) \\ (~x(x-1)~)^2 &=& (~(x-1)(x+1)~)^2 & (~(ab)^2 = a^2b^2~) \\ x^2(x-1)^2 &=& (x-1)^2(x+1)^2\\ x^2(x-1)^2 - (x-1)^2(x+1)^2 &=& 0\\ (x-1)^2 [ x^2 - (x+1)^2 ] &=& 0\\ (x-1)^2 ( x^2 - x^2-2x-1 ) &=& 0\\ (x-1)^2 ( -2x-1 ) &=& 0\\ (x-1)\cdot (x-1)\cdot ( -2x-1 ) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & -2x-1 = 0 && x = -\frac{1}{2} \end{array}\)
+130511 These are always a little difficult to wrap one's head around.....!!!
They are all solved in pretty much the same manner ....!!!
1. abs (x^2 - 1) = abs (x - 1)
We have two equations that result
x^2 - 1 = x - 1 and x^2 -1 = -(x - 1) working on both, we have
x^2 = x x^2 - 1 = -x + 1
x^2 - x = 0 x^2 + x - 2 = 0
x(x-1) = 0 (x +2) ( x-1) = 0
So x = 0 , 1 So x = -2 and x = 1
The solutions are x = -2, x = 0 and x = 1
Look at the graph of the solutions, here.........https://www.desmos.com/calculator/sanncuv3uw
+130511 2. 3|x-2|=2|x-3| divide both sides by 2
(3/2)|x-2|=lx-3| as before, we have.....
(3/2)(x - 2) = x -3 and (3/2)(x - 2) = -(x - 3)
(3/2)lx - 2 l = - x + 3
Multiply through by 2 in both equations.....
3(x -2) = 2x - 6
3x - 6 = 4x - 6 3(x - 2) = -2x + 6
The only solution to this is 3x - 6 = -2x + 6
when x = 0
5x = 12
x = 12/5 = 2 + 2/5 = 2.4
So the two solutions are x = 0 and x = 2.4
See the solution graphs, here..... https://www.desmos.com/calculator/frf6guc36z
+130511 3. |4-x|=5 this one is easy
Either
4 - x = 5 or 4 - x = -5
-x = -1 -x = -9
x = 1 x = 9
These solutions are easily verified in the original problem
+130511 4 . |2x-7|=|x-5|
2x - 7 = x - 5 or 2x - 7 = -(x - 5)
x = 2 2x - 7 = -x + 5
3x = 12
x = 4
Again......these are easily verified
+130511 5. |x^2-x|=|x^2-1| similar to the first one, we have
x^2 - x = x^2 - 1 or x^2 - x = - (x^2 - 1)
-x = -1 x^2 - x = -x^2 + 1
x = 1 2x^2 - x - 1 = 0
(2x +1) ( x - 1) = 0
Set each factor to 0
x = -1/2 [ x = 1 .....a repeated solution]
Here's the graph......https://www.desmos.com/calculator/kcbcpd3khj
+26400 1. |x2-1|=|x-1|
\(\begin{array}{rcll} |x^2-1| &=& |x-1| \qquad & (x^2-1) = (x-1)(x+1) \\ (|(x-1)(x+1)|)^2 &=& (|x-1|)^2 \qquad & (\text{square both sides}) \\ (~(x-1)(x+1)~)^2 &=& (x-1)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ (x-1)^2(x+1)^2 &=& (x-1)^2 \\ (x-1)^2(x+1)^2-(x-1)^2 &=& 0\\ (x-1)^2\left[(x+1)^2-1 \right] &=& 0\\ (x-1)^2(x^2+2x+1-1) &=& 0\\ (x-1)^2(x^2+2x) &=& 0\\ (x-1)^2\cdot x \cdot (x+2) &=& 0\\ (x-1)\cdot (x-1)\cdot x \cdot (x+2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & x = 0 \\ 3. & x+2 = 0 && x = -2 \end{array}\)
2. 3|x-2|=2|x-3|
\(\begin{array}{rcll} 3|x-2| &=& 2|x-3| \\ (3|x-2|)^2 &=& (2|x-3|)^2 \qquad & (\text{square both sides}) \\ 3^2(x-2)^2 &=& 2^2(x-3)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ 9(x-2)^2 &=& 4(x-3)^2 \\ 9(x^2-4x+4) &=& 4(x^2-6x+9) \\ 9x^2-36x+36 &=& 4x^2-24x+36 \\ 9x^2-36x &=& 4x^2-24x\\ 5x^2-12x &=& 0\\ x\cdot(5x-12) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 0 \\ 2. & 5x-12=0 && x=\frac{12}{5} \end{array}\)
3. |4-x|=5
\(\begin{array}{rcll} |4-x| &=& 5 \\ (|4-x|)^2 &=& (5)^2 \qquad & (\text{square both sides}) \\ (4-x)^2 &=& 25 \qquad & (\pm\sqrt{}) \\ 4-x &=& \pm 5\\ x &=& 4 \pm 5 \\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 4+5 && x = 9 \\ 2. & x = 4-5 && x = -1 \end{array}\)
4. |2x-7|=|x-5|
\(\begin{array}{rcll} |2x-7| &=& |x-5| \\ (|2x-7|)^2 &=& (|x-5|)^2 \qquad & (\text{square both sides}) \\ (2x-7)^2 &=& (x-5)^2 \\ 4x^2-28x+49 &=& x^2-10x+25\\ 3x^2-18x+24 &=& 0\qquad & :3\\ x^2-6x+8 &=& 0\\ (x-4)(x-2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-4=0 && x = 4 \\ 2. & x-2=0 && x=2 \end{array}\)
5. |x2-x|=|x2-1|
\(\begin{array}{rcll} |x^2-x| &=& |x^2-1| \\\\ \qquad x^2-x = x(x-1) \\ \qquad (x^2-1) = (x-1)(x+1) \\\\ |x(x-1)| &=& |(x-1)(x+1)| \\ (~|x(x-1)|~)^2 &=& (~|(x-1)(x+1)|~)^2 & (\text{square both sides}) \\ (~x(x-1)~)^2 &=& (~(x-1)(x+1)~)^2 & (~(ab)^2 = a^2b^2~) \\ x^2(x-1)^2 &=& (x-1)^2(x+1)^2\\ x^2(x-1)^2 - (x-1)^2(x+1)^2 &=& 0\\ (x-1)^2 [ x^2 - (x+1)^2 ] &=& 0\\ (x-1)^2 ( x^2 - x^2-2x-1 ) &=& 0\\ (x-1)^2 ( -2x-1 ) &=& 0\\ (x-1)\cdot (x-1)\cdot ( -2x-1 ) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & -2x-1 = 0 && x = -\frac{1}{2} \end{array}\)
