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Please help 50 points

Please help 50 points-example-1
User Iveytron
by
8.4k points

2 Answers

4 votes

Answer:

x-3

Explanation:


\sqrt{x^2 -x+9\\

First rewrite 9 as 3^2 so...
√(x^2 -6x+3^2)

Next multiply the equation by 2ab so... 2ab = 2 · x · -3

Simplify so... 2ab = -6x

Then use the Perfect Square Trinomial rule: a^2 - 2ab + b^2 = (a-b)^2

so.. a = x and b = -3 to look like this:
\sqrt(x- 3^2)

And then finally pull out terms under the radical to get...

x-3

User Quangdao Nguyen
by
7.8k points
4 votes

Answer:

0 < y <= 6

Explanation:

Given:

√(x² - 6x + 9)

if -3 <= x < 3

you can write (x² - 6x + 9)

as (x-3) * (x- 3) which is (x-3)²

so now you have √{(x-3)²}, which is the same as the absolute value of x-3. In mathematics it is written as:

| x - 3 |

So √(x² - 6x + 9) = | x - 3 |

See the graph in the attachment.

for -3 <= x < 3

you get the y values bigger then ( but not equal to) 0 and smaller then or equal to 6.

This is written as 0 < y <= 6

Please help 50 points-example-1
User Dina Diagovic
by
7.7k points

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