Question

could someone help me quick

Which of the following values "completes the square," or creates a perfect square trinomial, for x2 + 10x + ___?

25
10
5
–5

Answer 1

hint:
what is \((x+5)^2=...?\)

Answer 2

add square of half the coefficient of x

Answer 3

thank you @Michele_Laino could you help with another one by chance

Answer 4

ok!

Answer 5

Rewrite f(x) = –2(x − 3)2 + 2 from vertex form to standard form.

f(x) = –2x2 − 18
f(x) = –2x2 + 12x + 20
f(x) = –2x2 + 12x − 16
f(x) = 4x2 − 24x + 38

Answer 6

@Michele_Laino ^

Answer 7

here we have to compute the square of binomial, so we can write this:
\[f\left( x \right) = - 2\left( {{x^2} + 9 - 6x} \right) + 2 = ...?\]
please continue

Answer 8

so i got
f(x)= -2x^2 -18 +12x +2 ? is that right

Answer 9

@Michele_Laino

Answer 10

yes! Now you have to simplify again, so what is \(-18+2=...?\)

Answer 11

-16 so it would be answer choice b?
@Michele_Laino

Answer 12

you have to search for the option which contains \(-16\)

Answer 13

thats what i did. sorry to ask but do you have time for one more by chance @Michele_Laino

Answer 14

I think it is option C.
Yes! I can help you!

Answer 15

@Michele_Laino that what i meant i typed the wrong letter

Solve x2 + 8x − 3 = 0 using the completing-the-square method.

Answer 16

If I add and subtract 16, I get this:
\[\Large \begin{gathered}
{x^2} + 8x - 3 = {x^2} + 8x + 16 - 16 - 3 = \hfill \\
\hfill \\
= \left( {{x^2} + 8x + 16} \right) - 16 - 3 = ...? \hfill \\
\end{gathered} \]

Answer 17

I am very confused on this where did you get the 16 from @Michele_Laino

Answer 18

since the three terms inside the parentheses are the square of a binomial. What is such binomial?

Answer 19

@Michele_Laino i am sorry but i still am not getting this

Answer 20

please compute the square of this binomial:
\((x+4)^2=...?\)

Answer 21

i got x^2 + 8x + 16
thats where you got the 16 right? @Michele_Laino

Answer 22

yes! That is why I chose \(16\)

Answer 23

ok so so how do i solve this though @Michele_Laino

Answer 24

so we can rewrite your original equation as below:
\[\begin{gathered}
{x^2} + 8x - 3 = {x^2} + 8x + 16 - 16 - 3 = \hfill \\
\hfill \\
= \left( {{x^2} + 8x + 16} \right) - 16 - 3 = \hfill \\
\hfill \\
= {\left( {x + 4} \right)^2} - 19 \hfill \\
\end{gathered} \]
therefore:
\[{x^2} + 8x - 3 = 0 \Rightarrow {\left( {x + 4} \right)^2} - 19 = 0\]

Answer 25

In other words the original equation, is equivalent to this one:
\[{\left( {x + 4} \right)^2} - 19 = 0\]

Answer 26

my answer choices have an addition sign over a subtraction sign so what would the answer be @Michele_Laino

Answer 27

If we add \(19\) to both sides, we get:
\[{\left( {x + 4} \right)^2} - 19 + 19 = 0 + 19\]
please combine similar terms

Answer 28

ok now i understand i know i sound needy but is there anytime for 1 more i just dont understand it @Michele_Laino

Answer 29

ok! I can help you :)

Answer 30

Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

f(x) = (x − 8)2
f(x) = (x − 4)2 − 13
f(x) = (x − 4)2 + 3
f(x) = (x − 4)2 + 16

Answer 31

@MissViolet

Answer 32

@Michele_Laino

Answer 33

again I add and subtract \(16\) at the right side, so I get this:
\[\begin{gathered}
f\left( x \right) = {x^2} - 8x + 3 = {x^2} - 8x + 16 - 16 + 3 = \hfill \\
\hfill \\
= \left( {{x^2} - 8x + 16} \right) - 16 + 3 = \hfill \\
\end{gathered} \]

Answer 34

the three terms inside the parentheses are the square of a binomial. Do you know what is such binomial?

Answer 35

no i dont @Michele_Laino

Answer 36

please, compute this:
\[{\left( {x - 4} \right)^2} = ...?\]

Answer 37

x^2 - 8x + 16 ?

Answer 38

@Michele_Laino

Answer 39

correct!
so we can rewrite your function as below:
\[\begin{gathered}
f\left( x \right) = {x^2} - 8x + 3 = {x^2} - 8x + 16 - 16 + 3 = \hfill \\
\hfill \\
= \left( {{x^2} - 8x + 16} \right) - 16 + 3 = \hfill \\
\hfill \\
= {\left( {x - 4} \right)^2} - 16 + 3 \hfill \\
\end{gathered} \]
so, what is the right option?

Answer 40

Answer choice b
@Michele_Laino

Answer 41

correct!

Answer 42

thank you so much @Michele_Laino next time i need help ill come to you again

Answer 43

ok! :)

Answer 44

btw for anyone going to use this they are all correct