Question

find the roots of the quadratic equation
$\sqrt{ {3x }^{2} } - 2 \sqrt{2x} - 2 \sqrt{3} = 0$
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Answer 1

$\dag \: \underline{\sf AnsWer :}$

• We are given a quadratic equation as √3x² - 2√2x - 2√3 = 0 and we need to find the roots of the given quadratic equation. To find the roots the given quadratic equation we will use the method of splitting the middle term. So, let's solve :

$:\implies \sf \sqrt{3} {x}^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$

$:\implies \sf \sqrt{3} {x}^{2} - 3 \sqrt{2} x + \sqrt{2}x - 2 \sqrt{3} = 0$

$:\implies \sf \sqrt{3} x(x - \sqrt{6} ) + \sqrt{2} (x - \sqrt{6}) = 0$

$:\implies \sf (\sqrt{3}x+ \sqrt{2} )(x - \sqrt{6}) = 0$

$:\implies \underline{ \boxed{ \sf x = \frac{ - \sqrt{2} }{ \sqrt{3} } \: or \: x = \sqrt{6} }}$

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$\boxed{\boxed{\begin{minipage}{6.6 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}}$