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Which equation shows x^2+6x-2 = 0 rewritten by completing the square?

A. (x+3)^2=18
B. (x+3)^2-11
C. (x+3)^2-2
D. (X+3)^2-9

Sagot :

facundo3141592

Answer:

Option B.

Step-by-step explanation:

We have the equation:

x^2 + 6*x - 2 = 0

Remember that:

(a*x + b)^2 = (a*x)^2 + 2*a*b*x + b^2

Because the term with x^2 has a coefficient equal to 1, then a = 1.

(x + b)^2 = x^2 + 2*b*x + b^2

The ter with x is: 6*x

then:

2*b*x = 6*X

2*b = 6

b = 6/2 = 3

we have:

(x + 3)^2 = x^2 + 6*x + 9

Now, our equation is:

x^2 + 6*x - 2 = 0

Now we can add the term (11 - 11) = 0

x^2 + 6*x - 2 + (11 - 11) = 0

x^2 + 6*x - 2 + 11 - 11 = 0

x^2 + 6*x + (11 - 2) - 11 = 0

x^2 + 6*x + 9 - 11 = 0

And we know that:

x^2 + 6*x + 9  = (x + 3)^2

Then we can rewrite our equation as:

(x + 3)^2 - 11 = 0

Then the correct option is B.

Answer:

See below attachment

Step-by-step explanation:

A p E x

View image koolkieranryan
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