1.What are the zeros of the polynomial function? f(x)=x^2+9x+20 2.What are the zeros of the polynomial function? f(x)=x^2−4x−60 3.What are the roots of the equa
Mathematics
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Question
1.What are the zeros of the polynomial function?
f(x)=x^2+9x+20
2.What are the zeros of the polynomial function?
f(x)=x^2−4x−60
3.What are the roots of the equation?
x^2+24=14x
4.What are the zeros of the function?
g(x)=x^2−x−72
f(x)=x^2+9x+20
2.What are the zeros of the polynomial function?
f(x)=x^2−4x−60
3.What are the roots of the equation?
x^2+24=14x
4.What are the zeros of the function?
g(x)=x^2−x−72
1 Answer

1. User Answers David1993
Let's to the first example:
f(x) = x^2 + 9x + 20
Ussing the formula of basckara
a = 1
b = 9
c = 20
Delta = b^2  4ac
Delta = 9^2  4.(1).(20)
Delta = 81  80
Delta = 1
x = [ b +/ √(Delta) ]/2a
Replacing the data:
x = [ 9 +/ √1 ]/2
x' = (9 1)/2 <=>  5
Or
x" = (9+1)/2 <=>  4
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Already the second example:
f(x) = x^2 4x 60
Ussing the formula of basckara again
a = 1
b = 4
c = 60
Delta = b^2 4ac
Delta = (4)^2 4.(1).(60)
Delta = 16 + 240
Delta = 256
Then, following:
x = [ b +/ √(Delta)]/2a
Replacing the information
x = [ (4) +/ √256 ]/2
x = [ 4 +/ 16]/2
x' = (416)/2 <=> 6
Or
x" = (4+16)/2 <=> 10
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Now we are going to the 3 example
x^2 + 24 = 14x
Isolating 14x , but changing the sinal positive to negative
x^2  14x + 24 = 0
Now we can to apply the formula of basckara
a = 1
b = 14
c = 24
Delta = b^2 4ac
Delta = (14)^2 4.(1).(24)
Delta = 196  96
Delta = 100
Then we stayed with:
x = [ b +/ √Delta ]/2a
x = [ (14) +/ √100 ]/2
We wiil have two possibilities
x' = ( 14 10)/2 <=> 2
Or
x" = (14 +10)/2 <=> 12
________________
To the last example will be the same thing.
f(x) = x^2  x 72
a = 1
b = 1
c = 72
Delta = b^2 4ac
Delta = (1)^2 4(1).(72)
Delta = 1 + 288
Delta = 289
Then we are going to stay:
x = [ b +/ √Delta]/2a
x = [ (1) +/ √289]/2
x = ( 1 +/ 17)/2
We will have two roots
That's :
x = (1  17)/2 <=> 8
Or
x = (1+17)/2 <=> 9
Well, this would be your answers.