Question

√3x^2 -2√2x-2√3=0 solve by quadratic formula which is b^2-4ac

Answer 1

Heya !!

The given equation is √3X² - 2√2X - 2√3 = 0

Here,

a = √3 , b = -2√2 and c = -2√3.

Discriminant ( D ) = B²-4AC

=> (-2√2)² - 4 × √3 × -2√3

=> 8 + 24

=> 32

✓D = √32 = √2 × 2 × 2 × 2 × 2 = 4√2.

Roots of the given equation are : -B + √D/2A and -B - √D/2A

=> - (-2✓2) + 4√2 / 2√3 and -(-2√2) - 4√2/2√3

=> ( 2√2 + 4√2 ) /2√3 and (2√2 - 4√2 ) /2√3

=> ( 6√2/2√3 ) And ( -2√2/2√3 ).



=> ( 3√2/√3 ) and ( -√2/√3)



=> ( √3 × √3 × √2/✓3) and ( -√2/√3)



=> ( √3 × √2 ) and ( -√2/✓3)



=> ( √6 ) and ( -√2/√3).

Answer 2

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

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

$= > a = \sqrt{3},b = -2\sqrt{2},c = -2\sqrt{3}$

(1)

$x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{-(-2\sqrt{2})+\sqrt{(-2\sqrt{2})^2 - 4\sqrt{3}(-2\sqrt{3})}}{2\sqrt{3}}$

$= > \frac{2\sqrt{2}\sqrt{(2\sqrt{2})^2 + 4\sqrt{3} * 2\sqrt{3}}}{2\sqrt{3}}$

$= > \frac{2\sqrt{2}+\sqrt{(2\sqrt{2})^2 + 24}}{2\sqrt{3}}$

$= > \frac{2\sqrt{2} + \sqrt{32}}{2\sqrt{3}}$

$= > \frac{2\sqrt{2} + 4\sqrt{2}}{2\sqrt{3}}$

$= > \frac{6\sqrt{2}}{2\sqrt{3}}$

$= > \frac{3\sqrt{2}}{\sqrt{3}}$

$= > \sqrt{6}$

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(2)

$= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

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

$= >\frac{2\sqrt{2}-{4\sqrt{2}}}{2\sqrt{3}}$

$= > \frac{-2\sqrt{2}}{2\sqrt{3}}$

$= > -\sqrt{\frac{2}{3}}$

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Therefore the required quadratic solutions are|:

$= > x = \boxed {\sqrt{6}, -\sqrt{\frac{2}{3}} }$

Hope this helps