Question

Which function has real zeros at x = 3 and x = 7?

f(x) = x2 + 4x – 21
f(x) = x2 – 4x – 21
f(x) = x2 – 10x + 21
f(x) = x2 – 10x – 21

Answer 1

x = 3 x = 7 Okay well I got C. f(x) = x2 - 10x + 21 I got it because u subtract and add at the last part.

Answer 2

we know that

The zeros of the function are the values of x when the value of the function is equal to zero

so

Let's verify each case to determine the solution.

Substitute the value of
`x=3` and
`x=7` in each function and then verify the value of the function

case A)
`f(x)=x^(2) +4x-21`

For
`x=3`

`f(3)=3^(2) +4*3-21=0`

therefore

the value of
`x=3` is a zero of the function

For
`x=7`

`f(7)=7^(2) +4*7-21=56`

therefore

the value of
`x=7` is not a zero of the function

The case A) is not the solution

case B)
`f(x)=x^(2) -4x-21`

For
`x=3`

`f(3)=3^(2) -4*3-21=-24`

therefore

the value of
`x=3` is not a zero of the function

For
`x=7`

`f(7)=7^(2) -4*7-21=0`

therefore

the value of
`x=7` is a zero of the function

The case B) is not the solution

case C)
`f(x)=x^(2) -10x+21`

For
`x=3`

`f(3)=3^(2) -10*3+21=0`

therefore

the value of
`x=3` is a zero of the function

For
`x=7`

`f(7)=7^(2) -10*7+21=0`

therefore

the value of
`x=7` is a zero of the function

The case C) is a solution

case D)
`f(x)=x^(2) -10x-21`

For
`x=3`

`f(3)=3^(2) -10*3-21=-42`

therefore

the value of
`x=3` is not a zero of the function

For
`x=7`

`f(7)=7^(2) -10*7-21=-42`

therefore

the value of
`x=7` is not a zero of the function

The case D) is not a solution

therefore

The answer is

`f(x)=x^(2) -10x+21`