Question

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:​

Answer 1

Given:

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

To Find:

• Find the roots of each of the following equations

Solution:

On comparing with ax^2+bx+c =0,we get

Here,

$\sf \: a = \sqrt{3} \\ \sf b = - 2 \sqrt{2} \\ \sf \: c = - 2 \sqrt{3}$

$\sf \: (1) = x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a}$

Now put the values

$\sf \implies \: \frac{ - ( - 2 \sqrt{2} ) + \sqrt{ {( - 2 \sqrt{2)} }^{2} - 4 \sqrt{3}( - 2 \sqrt{3}) } }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{2 \sqrt{2} \sqrt{ {(2 \sqrt{2)} }^{2} + 4 \sqrt{3} \times 2 \sqrt{3} } }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{2 \sqrt{2} + \sqrt{ {(2 \sqrt{2} )}^{2} + 4 \sqrt{3} \times 2 \sqrt{3} } }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{2 \sqrt{2} + \sqrt{ {(2 \sqrt{2} )}^{2} + 24 } }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{2 \sqrt{2} + \sqrt{32} }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{2 \sqrt{2} + 4 \sqrt{2} }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{6 \sqrt{2} }{2 \sqrt{3} } \\ \\ \sf \implies \: \frac{3 \sqrt{2} }{ \sqrt{3} } \\ \\ \sf \implies \: x = 6$

$\sf \: (2)x = \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

Now put the values

$\sf \: x = \frac{ - ( - 2 \sqrt{2}) - \sqrt{ {(2 \sqrt{2}) }^{2} + 4 \sqrt{3} - 2 \sqrt{3} } }{2 \sqrt{3} } \\ \\ \sf \: x = \frac{2 \sqrt{2} - \sqrt{2} }{2 \sqrt{3} } \\ \\ \sf \: x = \frac{ - 2 \sqrt{2} }{2 \sqrt{3} } \\ \\ \sf \: x = - \sqrt{ \frac{2}{3} }$

Therefore x = √6,-√2/3.