what does resticted value mean?

• Jun 5th 2007, 06:42 PM
lostone
what does resticted value mean?

State the restricted values of x for 6x-5
2x(x+4)

I really am lost on this....:eek:

thanks
• Jun 5th 2007, 06:43 PM
ThePerfectHacker
Quote:

Originally Posted by lostone

State the restricted values of x for 6x-5
2x(x+4)

I really am lost on this....:eek:

thanks

It is when the denominator is zero:

\$\displaystyle 2x(x+4)=0\$

Make each factors zero and solve:
\$\displaystyle 2x=0 \mbox{ and }x+4=0\$
• Jun 5th 2007, 06:51 PM
lostone
??????
So that means the answer is zero?

Quote:

Originally Posted by ThePerfectHacker
It is when the denominator is zero:

\$\displaystyle 2x(x+4)=0\$

Make each factors zero and solve:
\$\displaystyle 2x=0 \mbox{ and }x+4=0\$

• Jun 5th 2007, 06:52 PM
ThePerfectHacker
Quote:

Originally Posted by lostone
So that means the answer is zero?

No!

We need to do both factors.

1)\$\displaystyle 2x=0 \Rightarrow x=0\$
2)\$\displaystyle x+4=0 \Rightarrow x=-4\$.

Thus,
\$\displaystyle x=0,-4\$ are the two restricted values.
• Jun 5th 2007, 07:05 PM
lostone
Why would the question be writen with only term "x" if it is used to make 2 different numbers = 0? Like y would it not b writen 6x-5
2x(y+4)

X= 0
y= -4 instead of "x" = 2 different numbers.
• Jun 5th 2007, 07:09 PM
ThePerfectHacker
Quote:

Originally Posted by lostone
Why would the question be writen with only term "x" if it is used to make 2 different numbers = 0? Like y would it not b writen 6x-5
2x(y+4)

X= 0
y= -4 instead of "x" = 2 different numbers.

There are two possible values for \$\displaystyle x\$.

For example,
Solve: \$\displaystyle x^2-1=0\$.

If you factor,
\$\displaystyle (x+1)(x-1)=0\$
Making factors equal to zero we find that,
\$\displaystyle x=-1\$ or \$\displaystyle x=1\$.

Those are the two possible values for x which solve this.

Same here, it does not mean that there has to be only one value for x there can be two (or more).
• Jun 5th 2007, 07:16 PM
lostone
so there is no rule that x has to equal 1 number....why would they write it that way? Just to make it more difficult?

Quote:

Originally Posted by ThePerfectHacker
No!

We need to do both factors.

1)\$\displaystyle 2x=0 \Rightarrow x=0\$
2)\$\displaystyle x+4=0 \Rightarrow x=-4\$.

Thus,
\$\displaystyle x=0,-4\$ are the two restricted values.

Quote:

Originally Posted by ThePerfectHacker
There are two possible values for \$\displaystyle x\$.

For example,
Solve: \$\displaystyle x^2-1=0\$.

If you factor,
\$\displaystyle (x+1)(x-1)=0\$
Making factors equal to zero we find that,
\$\displaystyle x=-1\$ or \$\displaystyle x=1\$.

Those are the two possible values for x which solve this.

Same here, it does not mean that there has to be only one value for x there can be two (or more).

• Jun 5th 2007, 07:26 PM
ThePerfectHacker
Quote:

Originally Posted by lostone
so there is no rule that x has to equal 1 number....why would they write it that way?

No there is not rule. Just write
\$\displaystyle x = 0\$ or \$\displaystyle x=-4\$.
That is it.
Quote:

Just to make it more difficult?
Stop complaining.
• Jun 5th 2007, 07:45 PM
lostone
I really appreciate the help!!! and I'm not complaining just having a hard time that's all.....Thank-you again!!!

Quote:

Originally Posted by ThePerfectHacker
No there is not rule. Just write
\$\displaystyle x = 0\$ or \$\displaystyle x=-4\$.
That is it.

Stop complaining.