Question

for what real numbers is x is x^2-6x+9 negative?
A. -33 C. x=-3 or x=3 D. 0< x<6 E. for no real number x

Answer 1

\[x^2-6x+9=\left( x-3 \right)^2 \ge 0\]

Answer 2

@surjithayer how does it go to (x-3)^2> 0

Answer 3

\[x^2-2*x*3+3^2=\left( x-3 \right)^2\]
square of a real number is always positive.

Answer 4

\[\left( a-b \right)^2=a^2-2ab+b^2\]

Answer 5

is b?

Answer 6

B

Answer 7

B is not correct. And, a reason for that choice should be given.

Answer 8

what do you mean by that @Directrix

Answer 9

Two Things:

The answer to this question is not B.

OpenStudy values the Learning process - not the ‘Give you an answer’ process

Don’t post only answers - guide the asker to a solution.

http://openstudy.com/code-of-conduct

Answer 10

First Factor it to:\[(x-3)(x-3)\] then from here we can analyze the signs of the parts:

Answer 11

Now guess what happens when we multiply those one either side of 3

Answer 12

those one either side of 3. Take a point, and multiply the signs on either side and see what you get!

Answer 13

for what real numbers is x is x^2-6x+9 **negative**?

Answer 14

so we multiply it by its self?

Answer 15

Name a value of x for which (x - 3) ^2 is negative.

Answer 16

whats a real; number?

Answer 17

Name any number that makes this negative: (x - 3) ^2

Answer 18

Read about real numbers here:

http://www.mathsisfun.com/numbers/real-numbers.html

Answer 19

another (x-3)

Answer 20

If you square any real number, the result will never be negative.

Answer 21

true?

Answer 22

Here is the incorrect option B: B. x<-3 or x>3

It says that if I pick any number greater than 3, then (x - 3) ^2 will be negative.

Let us test that.

(x - 3) ^2

x = 4

(4 - 3) ^2 = 1^2 = 1 which is POSITIVE and not negative.

Option B is wrong.

Answer 23

If you square a real number you will NEVER get a negative product.

Answer 24

Ok thanks