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Vanyuwa [196]
3 years ago
7

F(x) = 4x4-3x3–2x2-5 possible zeros

Mathematics
1 answer:
ddd [48]3 years ago
6 0

Answer:

f(x)= 1 correct me if wrong :)

Step-by-step explanation:

4 times 4 is 18

3 times 3 is 9

18-9=8

2 times 2 is 4

8-4=4

4-5=1

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Which are the solutions of x2 = 19x + 1?
Finger [1]

Answer:

(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})

Step-by-step explanation:

we have

x^2=19x+1

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2}-19x-1=0  

so

a=1\\b=-19\\c=-1

substitute in the formula

x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}

x=\frac{19\pm\sqrt{365}} {2}

x=\frac{19+\sqrt{365}} {2}

x=\frac{19-\sqrt{365}} {2}

(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})

therefore

StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction

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3 years ago
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