Question

What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?

© x=-2+, orx=-2-
• x=2 +, or x = 2-F
© x=-2+, or x=-2-F
• x=2+2 orx=2-7

Answer 1

x1 = 3.95

x2= 0.0425

Explanation:

Hi, to answer this question we have to apply the quadratic formula:

For: ax2+ bx + c

x =[ -b ± √b²-4ac] /2a

Replacing with the values given:

x =[ -(-24) ± √(-24)²-4(6)1] /2(6)

x = [ 24 ± √576 -24] /12

x = [ 24 ± √552] /12

x = [ 24 ± 23.49] /12

Positive:

x = [ 24 + 23.49] /12 = 47.49 /12 = 3.95

Negative:

x = [ 24 - 23.49] /12 = 0.51 /12 = 0.0425

Feel free to ask for more if needed or if you did not understand something.

Answer 2

`x_1 \approx 3.958`

`x_(2) \approx 0.042`

`f(x) = (x - 3.958)\cdot (x-0.042)`

Explanation:

The zeros of the quadratic function are obtained after some algebraic manipulation:

`f(x) = 6\cdot x^(2) - 24\cdot x + 1`

The roots of the second order polynomial are, respectively:

`x_(1,2) = (24 \pm √(576-4\cdot (6)\cdot (1)))/(2\cdot (6))`

`x_(1,2) \approx 2 \pm 1.958`

`x_1 \approx 3.958`

`x_(2) \approx 0.042`

`f(x) = (x - 3.958)\cdot (x-0.042)`