Find the zeros of f(x) = x^2 + 7x + 9

-7/2 ±1/2sqrt(13) = x

Step-by-step explanation:

f(x) =x^2 + 7x + 9

To find the zeros, set this equal to zero

0 = x^2 + 7x + 9

I will complete the square

Subtract 9 from each side

0-9 = x^2 + 7x + 9-9

-9 =x^2 + 7x

Take the coefficient of the x term, 7

divide by 2, 7/2

Then square it, (7/2)^2 = 49/4

Add this to both sides

-9 +49/4=x^2 + 7x + 49/4

-36/4 +49/4 = (x+7/2)^2

13/4 =  (x+7/2)^2

Take the square root of each side

±sqrt(13/4) = sqrt( (x+7/2)^2)

± sqrt(13) /sqrt(4)= (x+7/2)

± 1/2 sqrt(13) = (x+7/2)

Subtract 7/2 from each side

-7/2 ±1/2sqrt(13) = x+7/2-7/2

-7/2 ±1/2sqrt(13) = x

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl