Question

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Answer 1

Answer:

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$

Step-by-step explanation:

$2x^2 - 10x - 3 = 0 \\\\a = 2 \ , b = - 10 \ , \ c = - 3 \\\\x = \frac{-b^2\ \pm \ \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{10 \ \pm \sqrt{(-10)^2 - ( 4 \times 2 \times -3)} }{2 \times 2}\\\\x = \frac{10 \ \pm \sqrt{(100 - ( -24 )} }{4}\\\\x = \frac{10 \ \pm \sqrt{(100 + 24 } }{4}\\\\x = \frac{ 10 \ \pm \sqrt{124}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{4 \times 31}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{2^2 \times 31}}{4}\\\\x = \frac{ 10 \ \pm2 \sqrt{31}}{4}\\\\x = \frac{ 5 \ \pm\sqrt{31}}{2}$

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$