Question

Find the nature of the roots of the following quadratic equations,
(ii) x^2 -2 (√ 3)x - 1 = 0,

If real roots exist, find them,
hey plz any one answer it.​

Answer 1

Answer :-

→ The equation has real roots.

→ Roots are (√3 + 2) and (√3 - 2) .

Explanation :-

Given equation is :-

x² - 2√3x - 1 = 0

On comparing it with ax² + bx + c, we get :-

a = 1, b = -2√3, c = -1

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Firstly, let's check the nature of roots of the given equation.

D = b² - 4ac

⇒ D = (-2√3)² - 4(1)(-1)

⇒ D = 12 + 4

⇒ D = 16

As D > 0, so the equation has real and distinct roots.

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Now, let's calculate the roots of the equation using 'quadratic formula' / 'Shreedharacharya rule' :-

x = (-b ± √D)/2a

⇒ x = [-(-2√3) ± √16]/2(1)

⇒ x = [2√3 ± 4]/2

⇒ x = [2(√3 ± 2)]/2

⇒ x = √3 ± 2

⇒ x = √3 + 2 ; x = √3 - 2

Answer 2

Answer:

Given :-

• x² - 2(√3)x - 1 = 0

To Find :-

• What are the nature of the roots.

Formula Used :-

$\clubsuit$ Discriminate Formula :

$\mapsto \sf\boxed{\bold{\pink{Discriminate\: (D) =\: b^2 - 4ac}}}$

$\clubsuit$ Sridhara Acharya's Formula :

$\mapsto \sf\boxed{\bold{\pink{x =\: \dfrac{- b \pm \sqrt{b^2 - 4ac}}{2a}}}}$

Solution :-

First, we have to find the nature of the roots :

Given Equation :

$\leadsto \sf\bold{x^2 - 2(\sqrt{3})x - 1 =\: 0}$

where,

• a = 1
• b = - 2√3
• c = - 1

According to the question by using the formula we get,

$\implies \sf Discriminate\: (D) =\: (- 2\sqrt{3})^2 - 4(1)(- 1)$

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

$\implies \sf Discriminate\: (D) =\: 4 \times 3 + 4$

$\implies \sf Discriminate\: (D) =\: 12 + 4$

$\implies \sf\bold{\purple{Discriminate\: (D) =\: 16 > 0}}$

$\therefore$ The nature of the roots is real, unequal and distinct roots.

Now, we have to find the roots of the quadratic equation :

$\implies \sf x =\: \dfrac{- (- 2\sqrt{3}) \pm \sqrt{16}}{2(1)}$

$\implies \sf x =\: \dfrac{2\sqrt{3} \pm 4}{2}$

$\longrightarrow \sf x =\: \dfrac{2\sqrt{3} + 4}{2}$

$\longrightarrow \sf x =\: \dfrac{2(\sqrt{3} + 2)}{2}$

$\longrightarrow \sf x =\: \dfrac{\cancel{2}(\sqrt{3} + 2)}{\cancel{2}}$

$\longrightarrow \sf\bold{\red{x =\: \sqrt{3} + 2}}$

Either,

$\longrightarrow \sf x =\: \dfrac{2\sqrt{3} - 4}{2}$

$\longrightarrow \sf x =\: \dfrac{2(\sqrt{3} - 2)}{2}$

$\longrightarrow \sf x =\: \dfrac{\cancel{2}(\sqrt{3} - 2)}{\cancel{2}}$

$\longrightarrow \sf\bold{\red{x =\: \sqrt{3} - 2}}$

${\normalsize{\bold{\underline{\therefore\: The\: roots\: are\: \sqrt{3} + 2\: or\: \sqrt{3} - 2\: .}}}}$

EXTRA INFORMATION :-

❒ Quadratic Equation with one Variables :

✪ The general form of equation is ax² + bx + c.

[ Note : ◆ If a = 0, then the equation becomes to a linear equation.

◆ If b = 0, then the roots of the equation becomes equal but opposite in sign.

◆ If c = 0, then one of the roots is zero. ]