Question

Find the roots of the quadratic equation √3x²-2√2x-2√3=0​

Answer 1

Answer:See the answer below

Hope this will help you

Step-by-step explanation:

Answer 2

Answer:

$\large{\underline{\boxed{\sf x = \dfrac{3\sqrt{2}}{\sqrt{3}} \: \: or \: \: \dfrac{-\sqrt{2}}{\sqrt {3}}}}}$

Step-by-step explanation:

Given that;

√3x² - 2√2x - 2√3 = 0, on comparing the given equation with ax² + bx + c = 0,

• a = √3
• b = -2√2
• c = -2√3

Discriminant = b² - 4ac

⇒ D = (-2√2)² - 4 * √3 * (-2√3)

⇒ D = 8 + 24

⇒ D = 32

$\tt x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \tt x = \frac{ - ( - 2 \sqrt{2}) \pm \sqrt{32} }{2 \times \sqrt{3} } \\ \\ \tt x = \frac{2 \sqrt{2} \pm 4 \sqrt{2} }{2 \sqrt{3} }$

Therefore,

$\tt x = \frac{2 \sqrt{2} + 4 \sqrt{2} }{2 \sqrt{3} } \\ \\ \tt x = \frac{6 \sqrt{2} }{2 \sqrt{3} } \\ \\ \implies \bf x = \frac{3 \sqrt{2} }{ \sqrt{3} }$

Or,

$\tt x = \frac{2 \sqrt{2} - 4 \sqrt{2} }{2 \sqrt{3} } \\ \\ \tt x = \frac{ - 2 \sqrt{2} }{2 \sqrt{3} } \\ \\ \implies \bf x = \frac{ - \sqrt{2} }{ \sqrt{3} }$