Question

What are the zeros of the function?

h(w)=w^2+13w+42
Enter your answers in the boxes.

Answer 1

Here, h(x) = w² + 13w + 42

h(x) = w² + 6w + 7w + 42 = 0
w(w + 6) 7(w + 6) = 0

(w+6)(w+7) = 0
w = -6 or -7

In short, Your roots would be: -6 & -7

Hope this helps!

Answer 2

The zero's of the function are
`x=-6` and
`x=-7`

Explanation:

we have

`h(w)=w^(2) +13w+42`

we know that

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

so

equate the function to zero

`w^(2) +13w+42=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`w^(2) +13w+42=0`

so

`a=1\\b=13\\c=42`

substitute in the formula

`x=\frac{-13(+/-)\sqrt{13^(2)-4(1)(42)}} {2(1)}`

`x=\frac{-13(+/-)√(169-168)} {2}`

`x=\frac{-13(+/-)1} {2}`

`x=\frac{-13(+)1} {2}=-6`

`x=\frac{-13(-)1} {2}=-7`