egt1gi

2023-03-14

How to solve $\left|2+3x\right|=\left|4-2x\right|$?

pomorijujfo

Beginner2023-03-15Added 7 answers

Solutions to $\left|2+3x\right|=\left|4-2x\right|$ or equivalently

$\sqrt{{(2+3x)}^{2}}=\sqrt{{(4-2x)}^{2}}$ are integrated into the solutions for

$(2+3x)}^{2}={(4-2x)}^{2$ or

${(2+3x)}^{2}-{(4-2x)}^{2}=0$ or

$(2+3x+4-2x)(2+3x-4+2x)=0$

or

$(x+6)(5x-2)=0$

or $x=\{-6,\frac{2}{5}\}$ the two solutions are workable.

$\sqrt{{(2+3x)}^{2}}=\sqrt{{(4-2x)}^{2}}$ are integrated into the solutions for

$(2+3x)}^{2}={(4-2x)}^{2$ or

${(2+3x)}^{2}-{(4-2x)}^{2}=0$ or

$(2+3x+4-2x)(2+3x-4+2x)=0$

or

$(x+6)(5x-2)=0$

or $x=\{-6,\frac{2}{5}\}$ the two solutions are workable.

loise9oa

Beginner2023-03-16Added 4 answers

$\left|2+3x\right|=\left|4-2x\right|$

$(2+3x)}^{2}={(4-2x)}^{2$

$9{x}^{2}+12x+4=4{x}^{2}-16x+16$

$5{x}^{2}+28x-12=0$

$5{x}^{2}+30x-2x-12=0$

$5x\cdot (x+6)-2\cdot (x+6)=0$

$(5x-2)\cdot (x+6)=0$

Thus, ${x}_{1}=-6$ and $x}_{2}=\frac{2}{5$

$(2+3x)}^{2}={(4-2x)}^{2$

$9{x}^{2}+12x+4=4{x}^{2}-16x+16$

$5{x}^{2}+28x-12=0$

$5{x}^{2}+30x-2x-12=0$

$5x\cdot (x+6)-2\cdot (x+6)=0$

$(5x-2)\cdot (x+6)=0$

Thus, ${x}_{1}=-6$ and $x}_{2}=\frac{2}{5$

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