Answer: 10, 11, & 12
<u>Step-by-step explanation:</u>
Let x represent the age of the youngest child.
Their ages are consecutive so,
Youngest: x
Middle: x + 1
Oldest: x + 2
The age of the Youngest squared (x²) equals 8 times the Oldest [8(x + 2)] plus 4.
x² = 8(x + 2) + 4
x² = 8x + 16 + 4
x² = 8x + 20
x² - 8x - 20 = 0
(x - 10)(x + 2) = 0
x - 10 = 0 or x + 2 = 0
x = 10 or x = -2
Since age cannot be negative, x = -2 is not valid
So, the Youngest (x) is 10
the Middle (x + 1) is 11
and the Oldest (x + 2) is 12
Dude I don’t think using this for an exam is it but ok
Answer:
The area is 136
Step-by-step explanation:
7 x 16 = 112
1/2 x 8 x 6 = 24
24 + 112 = 136
Completing the square follows the principle of taking

and converting it into

where d is the 'correctional number' as I like to call it - i.e. the number that converts the expanded bracket into the +c, since the expanded bracket will give us

.
In this case, 2/2=1 so we have the first part:

.
Expanding this gives us

. We need c to be 9, so we can just add 8.
Putting this together:

Now we can solve it more easily.
Rearranging: