Question

Solve in R Ps : The vertical bars represent the absolute values

Answer 1

Given:

`\left|2x\text{ +10\mid-\mid x}^2\text{ +3x -10\mid -x+2=-3}\right?`

Step 1: Add |x^2+3x-10| to both sides.

`\begin{gathered} |2x+10|−|x^2+3x−10|−x+2+|x^2+3x−10|=−3+|x^2+3x−10| \\ |2x+10|−x+2=|x^2+3x−10|−3 \end{gathered}`

Step 2: Step 2: Add x to both sides.

`\begin{gathered} \left|2x+10|−x+2+x=|x2^+3x−10|−3+x\right? \\ |2x+10|+2=|x^2+3x−10|+x−3 \end{gathered}`

Step 3: Step 3: Add -2 to both sides.

`\begin{gathered} |2x+10|+2+−2=|x2+3x−10|+x−3+−2 \\ |2x+10|=|x2+3x−10|+x−5 \end{gathered}`
`Either\text{ }2x+10=|x^2+3x−10|+x−5\text{ }or\text{ }2x+10=−\left(|x^2+3x−10|+x−5\right)`
`Part1:2x+10=|x^2+3x−10|+x−5`

(Flip the equation)

`\begin{gathered} |x^2+3x−10|+x−5=2x+10 \\ |x^2+3x−10|+x−5+−x=2x+10+−x\left(Add-x\text{ }to\text{ }both\text{ }sides\right) \\ |x^2+3x−10|−5=x+10 \\ |x^2+3x−10|−5+5=x+10+5\left(Add\text{ }5\text{ }to\text{ }both\text{ }sides\right) \\ |x^2+3x−10|=x+15 \\ We\text{ }know\text{ }either\text{ }x^2+3x−10=x+15\text{ }or\text{ }x^2+3x−10=−\left(x+15\right) \end{gathered}`
`\begin{gathered} x^2+3x−10=x+15\left(Possibility1\right) \\ x^2+3x−10−\left(x+15\right)=x+15−\left(x+15\right)\left(Subtract\text{ }x+15\text{ }from\text{ }both\text{ }sides\right) \\ x^2+2x−25=0 \\ For\text{ }this\text{ }equation:a=1,b=2,c=-25 \\ 1x^2+2x+−25=0 \\ x=\text{ }(-b\pm√(b^2-4ac))/(2a) \\ x=−1+√26\text{ }or\text{ }x=−1−√26 \end{gathered}`
`\begin{gathered} x^2+3x−10=−\left(x+15\right)\left(Possibility\text{ }2\right) \\ x^2+3x−10=−x−15\left(Simplify\text{ }both\text{ }sides\text{ }of\text{ }the\text{ }equation\right) \\ x^2+4x+5=0 \\ x\text{ = }\frac{-4\text{ }\pm\text{ }√(-4)}{2} \end{gathered}`

Check answers. (Plug them in to make sure they work.)

`\begin{gathered} x=−1+√26\text{ }\lparen Works) \\ x=−1−√26\left(Doesn^{\prime\text{ }}twork\right) \end{gathered}`

`Part2:2x+10=−\left(|x^2+3x−10|+x−5\right)`

(Flip the equation)

`\begin{gathered} −|x^2+3x−10|−x+5=2x+10 \\ −|x^2+3x−10|−x+5+x=2x+10+x\left(Addx\text{ }to\text{ }both\text{ }sides\right) \\ −|x^2+3x−10|+5=3x+10 \\ −|x^2+3x−10|+5+−5=3x+10+−5\left(Add-5tobothsides\right) \\ |x^2+3x−10|=−3x−5\text{ } \\ We\text{ }know\text{ }either\text{ }x^2+3x−10=−3x−5\text{ }or\text{ }x^2+3x−10=−\left(−3x−5\right) \end{gathered}`
`\begin{gathered} x^2+3x−10=−3x−5\left(Possibility1\right) \\ Solving \\ x=−3+√14\text{ }or\text{ }x=−3−√14 \end{gathered}`

`\begin{gathered} x^2+3x−10=−\left(−3x−5\right)\left(Possibility2\right) \\ x=√15\text{ }o\text{ }rx=−√15 \end{gathered}`

Check answers:

`\begin{gathered} x=−3+√14\left(Doesn^(\prime)t\text{ }work\text{ }in\text{ }original\text{ }equation\right) \\ x=−3−√14\left(Works\text{ }in\text{ }original\text{ }equation\right) \\ x=√15\left(Doesn^(\prime)t\text{ }work\text{ }in\text{ }original\text{ }equation\right) \\ x=−√15\left(Doesn^(\prime)t\text{ }work\text{ }in\text{ }original\text{ }equation\right) \end{gathered}`

Answer:

`x=−1+√26\text{ }or\text{ }x=−3−√14`