Question

The quadratic equation whose roots are 3 ± $\sqrt{5}$ is

• A. X² - 6X - 4 = 0
• B. y² - 6y + 4 = 0
• C. z² + 6z + 4 = 0
• D. a² - 6a - 4 = 0

Answer 1

The correct option is B

y² - 6y + 4 = 0

$y^2$ -6y+4=0

using formula

$\frac{6\pm \sqrt{(36-16)}}{2}$

= $\frac{6\pm 2\sqrt{5}}{2}$

=3±$\sqrt{5}$