Respuesta :

xero099

Answer:

[tex]x_1 \approx 3.958[/tex]

[tex]x_{2} \approx 0.042[/tex]

[tex]f(x) = (x - 3.958)\cdot (x-0.042)[/tex]

Step-by-step explanation:

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

[tex]f(x) = 6\cdot x^{2} - 24\cdot x + 1[/tex]

The roots of the second order polynomial are, respectively:

[tex]x_{1,2} = \frac{24 \pm \sqrt{576-4\cdot (6)\cdot (1)}}{2\cdot (6)}[/tex]

[tex]x_{1,2} \approx 2 \pm 1.958[/tex]

[tex]x_1 \approx 3.958[/tex]

[tex]x_{2} \approx 0.042[/tex]

[tex]f(x) = (x - 3.958)\cdot (x-0.042)[/tex]

Answer:

x1 = 3.95

x2= 0.0425

Step-by-step 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.

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