This is a REALLY difficult question to answer (Atleast in my knowledge), but I will try my best:
Input:
Solve for x:
|3x-2|-1=|5-x|
Intepretation: Sove for \(x\) in the function \(|3x-2|-1=|5-x|\)
Step 1. Rewrite the expressions:
\(\sqrt{\left(3x-2\right)^2}-1=\sqrt{\left(5-x\right)^2}\)
Step 2. Square both sides:
\(({\sqrt{\left(3x-2\right)^2}-1})^2=(\sqrt{\left(5-x\right)^2})^2\)
Step 3. Expand x2:
\((3x-2)^2-2\sqrt{\left(3x-2\right)^2}+1=(5-x)^2\)
\(9x^4-12x+5-2\sqrt{\left(3x-2\right)^2}=x^2-10x+25\)
Step 4. Move the corresponding terms:
\(8x^2-2x-20=2\sqrt{\left(3x-2\right)^2}\)
Step 5. Square both sides again:
\((8x^2-2x-20)^2=(2\sqrt{\left(3x-2\right)^2})^2\)
\(64x^4-16x^3-160x^2-16x^3+4x^2+40x-160x^2+40x+400)=4(3x-2)^2\)
Step 6. Simplify:
\(64x^4-32x^3-316x^2+80x+400=36x^2-48x+16\)
Step 7. Move the terms again:
\(64x^4-32x^3-352x^2+128x+384=0\)
Step 8. Divide by their greatest common divisor (i.e. \(32\))
\(2x^4-x^3-11x^2+4x+12=0\)
Step 9. Factorize the function:
Now we use the rational root theorem:
For a polynomial \(ax^n+bx^{n-1}+cx^{n-2}+...+yx^1+z\)
The possible rational roots will be located at \(± \frac{all\space\space factors\space of\space a}{all\space factors\space of\space z}\)
Now find the factors of both \(a\) and \(z\)
9.1 The factors of \(12=z\) is: \(1,2,3,4,6,12\)
The factors of \(2=a\) is: \(1,2\)
9.2 Plug the factors in:
\(±\frac{1,2,3,4,6,12}{1,2}\)
9.3 Expand & List them out:
\(±1,2,3,4,6,12,1,\frac{3}{2},\frac{1}{2},2,3,6\)
9.4 Cancel & Reorganize the repeated possible roots:
\(±\frac{1}{2},1,2,\frac{3}{2},3,4,6,12\)
9.5 Plug all the numbers above into the equation
\(2x^4-x^3-11x^2+4x+12=0\)
Is it \(±\frac{1}{2}\)? No.
Is it \(±1\)? \(1\) doesn't, while \(-1\) does
We can deduct that \((x+1)\) is a factor of the polynomial:
9.6 Perform polynomial long division:
Solve for \(\frac{2x^4-x^3-11x^2+4x+12}{x+1}=(2x^3-3x^2-8x+12)\)
9.7 Factor \((2x^3-3x^2-8x+12)\)
Factor \(x^2\) from \(2x^3-3x^2\), \(-4\) from \(-8x+12\)
\(=x^2(2x-3)-4(2x-3)\)
Merge the terms:
\(=(x^2-4)(2x-3)\)
9.71 Factor \((x^2-4)\)
Use the law: \(a^2-b^2=(a+b)(a-b)\) with \(a=1\) and \(b=2\)
\(=(x+2)(x-2)\)
Therefore:
\(2x^4-x^3-11x^2+4x+12=(x+1)(x+2)(x-2)(2x-3)\)
Step 10. Find the possible roots of \(x\)
\(x+1=0, x+2=0, x-2=0, 2x-3=0\)
\(x=-1\space or\space -2\space or\space 2\space or\space \frac{3}{2}\)
Step 11. Final Step:
Plug the four \(x\) above into \(|3x-2|-1=|5-x|\)
And making sure the left and the right side is the same:
In the end: \(x=-2,2\)
Therefore the final solution for \(|3x-2|-1=|5-x|\) is:
\(x=±2\)
(There must be an easier way to solve this.)
