What is x if $3 x - \sqrt{2 {(x - 3)}^{2}} = 2 x + 7 - 5 x$ $3x - sqrt(2(x-3)^2) = 2x + 7-5x$ ?

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What is x if $3 x - \sqrt{2 {(x - 3)}^{2}} = 2 x + 7 - 5 x$ $3x - sqrt(2(x-3)^2) = 2x + 7-5x$ ?

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Answer 1 · 2018-08-04T12:52:17

$x_{1} = \frac{72 + 22 \sqrt{2}}{34 \cdot 2} = \frac{36 + 11 \sqrt{2}}{34}$ $x_1=(72+22sqrt2)/(34*2)=(36+11sqrt2)/34$ and $x_{2} = \frac{72 - 22 \sqrt{2}}{34 \cdot 2} = \frac{36 - 11 \sqrt{2}}{34}$ $x_2=(72-22sqrt2)/(34*2)=(36-11sqrt2)/34$

Explanation:

$3 x - \sqrt{2 \cdot {(x - 3)}^{2}} = 2 x + 7 - 5 x$ $3x-sqrt(2*(x-3)^2)=2x+7-5x$

$3 x - \sqrt{2 \cdot {(x - 3)}^{2}} = 7 - 3 x$ $3x-sqrt(2*(x-3)^2)=7-3x$

$3 x + 3 x - 7 = \sqrt{2 \cdot {(x - 3)}^{2}}$ $3x+3x-7=sqrt(2*(x-3)^2)$

$6 x - 7 = \sqrt{2 \cdot {(x - 3)}^{2}}$ $6x-7=sqrt(2*(x-3)^2)$

${(6 x - 7)}^{2} = 2 \cdot {(x - 3)}^{2}$ $(6x-7)^2=2*(x-3)^2$

$36 x^{2} - 84 x + 49 = 2 \cdot (x^{2} - 6 x + 9)$ $36x^2-84x+49=2*(x^2-6x+9)$

$36 x^{2} - 84 x + 49 = 2 x^{2} - 12 x + 18$ $36x^2-84x+49=2x^2-12x+18$

$34 x^{2} - 72 x + 31 = 0$ $34x^2-72x+31=0$

$Delta=(-72)^2-4*34*31=968=(22sqrt2)^2$

Hence $x_1=(72+22sqrt2)/(34*2)=(36+11sqrt2)/34$ and $x_2=(72-22sqrt2)/(34*2)=(36-11sqrt2)/34$

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What is x if $3 x - \sqrt{2 {(x - 3)}^{2}} = 2 x + 7 - 5 x$ $3x - sqrt(2(x-3)^2) = 2x + 7-5x$ ?

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What is x if $3 x - \sqrt{2 {(x - 3)}^{2}} = 2 x + 7 - 5 x$ $3x - sqrt(2(x-3)^2) = 2x + 7-5x$ ?