Question

ans it urgent.


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Answer 1

★Given :-

• The equation $\sf x^2+2\sqrt{3} x+3=0$

★To find :-

• The root of the given equation.

★Solution :-

Here we are given an equation to solve using the quadratic formula.With the help of the flowchart,we first compare the the given equation with general form of quadratic polynomial ax² + bx + c, where a, b, and c are real numbers, and a≠ 0.

Comparing the equation with $\sf ax^2+bx+c=0$,

• a = 1
• b = 2√3
• c = 3

~Value of b²-4ac :

$\sf \implies (2\sqrt{3}) ^{2} - 4(1)(3) \\\\\implies \sf 4(3) - 12 \\\\\implies \sf 12 - 12 = 0$

~Using the formula :

$\displaystyle \leadsto \underline{\boxed{\tt x = \frac{-b \pm \sqrt{b^2-4ac} }{2a} }}$

Substituting the values,

$\displaystyle \implies \sf \frac{-2\sqrt{3}\ \pm \sqrt{(2\sqrt{3})^2 - 12}}{2(1)} \\\\\implies \sf \dfrac{-2\sqrt{3}\ \pm \sqrt{0} }{2} \\\\\implies \sf \frac{-\not{2}\sqrt{3} }{\not{2}} \\\\\therefore \underline{\boxed{\bold{ x = -\sqrt{3} ,-\sqrt{3} }}}$

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Answer 2

Step-by-step explanation:

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