Question

( 1 ). √3²+ 2x - 8√3
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Answer 1

Step-by-step explanation:

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$Given : \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0Given:3x2−22x−23=0</p><p></p><p>On comparing with ax^2 + bx + c = 0, we get</p><p></p><p>= > a = \sqrt{3},b = -2\sqrt{2},c = -2\sqrt{3}=>a=3,b=−22,c=−23</p><p></p><p>(1)</p><p></p><p>x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}x=2a−b+b2−4ac</p><p></p><p>= > \frac{-(-2\sqrt{2})+\sqrt{(-2\sqrt{2})^2 - 4\sqrt{3}(-2\sqrt{3})}}{2\sqrt{3}}=>23−(−22)+(−22)2−43(−23)</p><p></p><p>= > \frac{2\sqrt{2}\sqrt{(2\sqrt{2})^2 + 4\sqrt{3} * 2\sqrt{3}}}{2\sqrt{3}}=>2322(22)2+43∗23</p><p></p><p>= > \frac{2\sqrt{2}+\sqrt{(2\sqrt{2})^2 + 24}}{2\sqrt{3}}=>2322+(22)2+24</p><p></p><p>= > \frac{2\sqrt{2} + \sqrt{32}}{2\sqrt{3}}=>2322+32</p><p></p><p>= > \frac{2\sqrt{2} + 4\sqrt{2}}{2\sqrt{3}}=>2322+42</p><p></p><p>= > \frac{6\sqrt{2}}{2\sqrt{3}}=>2362</p><p></p><p>= > \frac{3\sqrt{2}}{\sqrt{3}}=>332</p><p></p><p>= > \sqrt{6}=>6</p><p></p><p>---------------------------------------------------------------------------------------------------------------</p><p></p><p>(2)</p><p></p><p>= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}=>x=2a−b−b2−4ac</p><p></p><p>= > x = \frac{-(-2\sqrt{2})-\sqrt{(2\sqrt{2})^2+4\sqrt{3}*2\sqrt{3}}}{2\sqrt{3}}=>x=23−(−22)−(22)2+43∗23</p><p></p><p>= >\frac{2\sqrt{2}-{4\sqrt{2}}}{2\sqrt{3}}=>2322−42</p><p></p><p>= > \frac{-2\sqrt{2}}{2\sqrt{3}}=>23−22</p><p></p><p>= > -\sqrt{\frac{2}{3}}=>−32</p><p></p><p>---------------------------------------------------------------------------------------------------------------</p><p></p><p>Therefore the required quadratic solutions are|:</p><p></p><p>= > x = \boxed {\sqrt{6}, -\sqrt{\frac{2}{3}} }=>x=6,−32</p><p></p><p>Hope this helps</p><p></p><p>$

Answer 2

x=5.42 is the correct answer