Question

ғɪɴᴅ ᴛʜᴇ ɴᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ʀᴏᴏᴛs ᴏғ ᴛʜᴇ ғᴏʟʟᴏᴡɪɴɢ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴs, (ɪɪ) x^2 -2 (√ 3)x - 1 = 0, ɪғ ʀᴇᴀʟ ʀᴏᴏᴛs ᴇxɪsᴛ, ғɪɴᴅ ᴛʜᴇᴍ, ʜᴇʏ ᴘʟᴢ ᴀɴʏ ᴏɴᴇ ᴀɴsᴡᴇʀ ɪᴛ.​

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

$\huge\mathfrak\pink{Answer}$

→ 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

________________________________

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.

________________________________

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