Absolute Value (algebraic version)....1

AI Thread Summary
The discussion focuses on rewriting absolute value expressions without using absolute value notation. For the first question, the expression |1 - sqrt{2}| + 1 simplifies to sqrt{2}. In the second question, given that x < 3, the expression |x - 3| is rewritten as -x + 3. Participants express agreement on the solutions provided. The thread highlights the challenges of learning precalculus while successfully addressing absolute value concepts.
nycmathguy
Homework Statement
Rewrite each expression without using absolute value notation.
Relevant Equations
n/a
Absolute Value (algebraic version)
Rule:

| x | = x when x ≥ 0

| x | = -x when x > 0

Rewrite each expression without using absolute value notation.

Question 1

|1 - sqrt{2} | + 1

The value 1 - sqrt{2} = a negative value.

So, -(1 - sqrt{2}) = - 1 + sqrt{2}.

When I put it all together, I get this:

-1 + sqrt{2} + 1

Answer: sqrt{2}

You say?

Question 2

| x - 3 | given that x < 3.

If x < 3, then x - 3 is a negative value.

So, -(x - 3) becomes -x + 3.

You say?
 
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I agree also for this.
Ssnow
 
Ssnow said:
I agree also for this.
Ssnow
I got it right again. Not bad for a person that has been attacked since joining the site for trying to learn precalculus, a subject that IS WAY OVER MY HEAD.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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