Question

Solve the following quadratic equation by Formula method:
X^2-3√3 x + 6 = 0​

Answer 1

Solution:

Given equation,

$\tt {x}^{2} - 3 \sqrt{3} x + 6 = 0$

We have to find out the roots of the given equation by quadratic formula.

So, comparing with ax² + bx + c = 0, we get,

• a = 1
• b = -3√3
• c = 6

So,

$\tt \implies x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\tt \implies x = \dfrac{3 \sqrt{3} \pm \sqrt{ {( - 3 \sqrt{3} )}^{2} - 4 \times 1 \times 6 } }{2}$

$\tt \implies x = \dfrac{3 \sqrt{3} \pm \sqrt{27 -24 }}{2}$

$\tt \implies x = \dfrac{3 \sqrt{3} \pm \sqrt{3}}{2}$

Thus,

$\tt \implies x_{1}= \dfrac{3 \sqrt{3} + \sqrt{3}}{2} = \dfrac{4 \sqrt{3} }{2} = 2 \sqrt{3}$

$\tt \implies x_{2}= \dfrac{3 \sqrt{3} - \sqrt{3}}{2} = \dfrac{2 \sqrt{3} }{2} = \sqrt{3}$

★ So, the roots of the given quadratic equation are 2√3 and √3.

Answer:

• x = √3, 2√3

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