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

Aug 4, 2018

${x}_{1} = \frac{72 + 22 \sqrt{2}}{34 \cdot 2} = \frac{36 + 11 \sqrt{2}}{34}$ and ${x}_{2} = \frac{72 - 22 \sqrt{2}}{34 \cdot 2} = \frac{36 - 11 \sqrt{2}}{34}$

Explanation:

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

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

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

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

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

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

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

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

$\Delta = {\left(- 72\right)}^{2} - 4 \cdot 34 \cdot 31 = 968 = {\left(22 \sqrt{2}\right)}^{2}$

Hence ${x}_{1} = \frac{72 + 22 \sqrt{2}}{34 \cdot 2} = \frac{36 + 11 \sqrt{2}}{34}$ and ${x}_{2} = \frac{72 - 22 \sqrt{2}}{34 \cdot 2} = \frac{36 - 11 \sqrt{2}}{34}$