Math, asked by nobinthomas02, 4 months ago

find the zeroes of x²-root 3x-6 plz fast

Answers

Answered by ItzAnu2108
2

 {x}^{2}  -  \sqrt{3} x - 6 \\  {x}^{2}  - 2 \sqrt{3} x +  \sqrt{3} x - 6 \\ x(x - 2 \sqrt{3} ) +  \sqrt{3} (x - 2 \sqrt{3} ) \\ (x +  \sqrt{3} )(x - 2 \sqrt{3} )

u'r done!!

Thnx✌️♥️~

Answered by Anonymous
1

x² - √3x - 6

x² - 2√3x + √3x - 6

x(x - 2√3) + √3(x - 2√3)

(x - 2√3)(x + √3)

x = 2√3 or -√3

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Answered by Anonymous
1

x² - √3x - 6

x² - 2√3x + √3x - 6

x(x - 2√3) + √3(x - 2√3)

(x - 2√3)(x + √3)

x = 2√3 or -√3

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Similar questions