Question

→ Determine the nature of the roots for the quadratic equation
$\sqrt{3x {}^{2} } + \sqrt{2x} - 2 \sqrt{3} = 0$
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Answer 1

Given :

• √3x² + √2x - 2√3 = 0

To find

• Nature of roots

Solution :

• According to the quadratic formula

★ D = b² - 4ac ★

• Where " D " is known as discriminant

• a = √3
• b = √2
• c = - 2√3

→ D = (√2)² - 4 × √3 × (-2√3)

→ D = 2 + 4 × 2 × 3

→ D = 2 + 24

→ D = 26

So, the given equation has two distinct real roots.

There are three situations in a quadratic equation

• Quadratic equation = ax² + bc + c = 0, where a ≠ 0 and a, b, c are real numbers

1. If b² - 4ac > 0, then two distinct real roots
2. If b² - 4ac < 0, then no real roots
3. If b² - 4ac = 0, then two equal real roots.

Above solution follows the first situation

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Answer 2

Answer:

Required Answer :-

By, using Quardtic formula

$\large \sf \: D = {b}^{2} - 4ac$

Here,

D = ?

b = √2

a = √3

c = -2√3

$\sf \: D = { \sqrt{2} }^{2} - 4( \sqrt{3} )( - 2 \sqrt{3} )$

$\sf \: D = 2 - 4( \sqrt{3} )( - 2 \times \sqrt{3})$

$\sf \: D = 2 + 4 \times 2 \times 3$

$\sf \: D = 2 + 24$

$\sf \pink{D = 26}$

It has two distinct real roots as it is greater than 0