Please HELP: Using the completing-the-square method, rewrite f(x) = x^2 + 4x − 1 in vertex form. f(x) = (x + 2)^2 + 1 f(x) = (x + 2)^2 f(x) = (x + 2)^2 + 4 f(x) = (x + 2)^2 − 5

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Please HELP: Using the completing-the-square method, rewrite f(x) = x^2 + 4x − 1 in vertex form. f(x) = (x + 2)^2 + 1 f(x) = (x + 2)^2 f(x) = (x + 2)^2 + 4 f(x) = (x + 2)^2 − 5

Mathematics
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do you know what a "perfect square trinomial" is? sometimes just called a "perfect square"
yes
\[x^2+bx+c=(x+\frac{b}{2})^2+c-\frac{b^2}{4}\]

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Other answers:

would it be D then?
\[f(x) = x^2+4x-1\]\[=(x^2+4x)-1\]\[=(x^2+4x+\color{red}{4})-1\color{red}{-4}\]\[=\boxed{(x-2)^2-5}\]
so... \(\bf f(x)=x^2+4x-1\implies f(x)=(x^2+4x+{\color{red}{ \square ?}}^2)-1\) what do you think is the missing number?
Lovely.
correct @iwanttogotostanford
one more?
@jdoe0001 why isn't `\boxed` working for me?
Which of the following values "completes the square," or creates a perfect square trinomial, for x2 + 6x + ___? 1 3 6 9
i believe its 9 but i just want to make sure
\[c=\left(\frac{6}{2}\right)^2 = ~?\]
9!
\[\color{red}{\checkmark}\]
i really need help with this one real quick, thanks for all your help..:-)
Solve x2 + 8x − 3 = 0 using the completing-the-square method. x = four plus or minus the square root of three x = negative four plus or minus the square root of three x = four plus or minus the square root of nineteen x = negative four plus or minus the square root of nineteen
use the method we used above.
so , B?
I don't know,I didn't solve it.
could you help me? :_)
first add +3 to both sides.
what do you get, @iwanttogotostanford ?
the same but on the left -6 and on the right 3
??....
okay, if you add +3 to both sides , the answer i was looking for is, x\(^2\)+8x=3
@Jhannybean sorry, i was gone

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