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

Answer:

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

Step-by-step 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

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

$x=\frac{-13(+/-)\sqrt{169-168}} {2}$

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

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

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