Question

find the roots of the quadratic equation by quadratic formula root 3 x square minus 2 root 2 x minus 2 root 3 equal to zero​

Answer 1

Answer:

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

Step-by-step explanation:

Given equation;

√3x² - 2√2x - 2√3 = 0

On comparing the given equation with ax² + bx + c = 0, thus

• a = √3
• b = -2√2
• c = -2√3

Discriminant = b² - 4ac

⇒ D = (-2√2)² - 4 * √3 * (-2√3)

⇒ D = 8 + 24

⇒ D = 32

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

Thus,

$\boxed {\bf x = \dfrac{ \sqrt{2} + 2 \sqrt{2} }{ \sqrt{3} } \: or \: \dfrac{ \sqrt{2} - 2 \sqrt{2} }{ \sqrt{3} } }$