Melody

Find a polynomial q(x) such that (x+1)^3+x^2*q(x) has degree less than 2.

You just have to cancel out the x^3 term.

q(x)= -x+k    where k is a constant.  Any real value of k will work fine.

In the five-digit base ten numeral ABCDE, different letters do not necessasrily represent different digits.  If this numeral is the fourth power of an integer and if  A + C + E = B + D, find the digit C.

10^4=10000    

17^4=83521

19^4 is too big.  so there are only  17-10+1= 8 possible numbers

Since there is only 8 possibilities I would just list them and see which one works for the given equation.

This is what you should be doing :

\(|x^2-20x-1|=20\\~\\ \color{blue}x^2-20x-1=20 \qquad or \qquad x^2-20x-1=-20 \\\)

       Think about this simple example

       |x|=5

       What this says it that the distance x is from 0 is 5.  So x could be 5 or maybe -5.

If you let    \(y=x^2-20x-1\)

then your question is asking, for what values of x is   y  20 units away from 0.

So y can equal 20  or   y can equal -20.

You will get 4 answers.  They will not necessarily all be real.

I expect you are only interested in real answers so if some are not real you can discard them.

Now you must substitute the real ones into the original equation to check that they work properly.

Discard any that do not work.

Is that enough help ?

You are very welcome EpicWaters.

Yes, at some point in time the Americans decided to get rid of some of the illogical spelling in the English language.

So there are now two recognised subsets of written English.  The english one and the USA one.  I am Australian so the correct one for me to use is the English one.

To bad Americans can't extend their logic and get rid of their imperial units of measure.

Most other countries got rid of those around 40 years ago.

I answered this yesterday.

Look back at older questions to find it.

\(x=1+\dfrac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\\ x-1=\dfrac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\\ x-1=\dfrac{1}{1+\frac{1}{2+(x-1)}}\\ x-1=\dfrac{1}{\frac{2+x-1+1}{2+x-1}}\\ x-1=\dfrac{1}{\frac{2+x}{1+x}}\\ x-1=\dfrac{1+x}{x+2}\\ (x-1)(x+2)=x+1\\ x^2+x-2=x+1\\ x^2=3\\ x=\pm\sqrt3\)

The answer must be positive so the only valid andwer is     \(x=\sqrt3\)

Maybe the digits are not in that order......

Hi Tom,

I do these by colour coding the sides (horizonal and vertical) and imagining them be pushed out to get rid of all the indents.

So for me this problems becomes.

horizontal    10+7+7+10 =  34

vertical     12+12+ 3+3   = 30

altogether = 64

So I believe your answer is correct.

6+4-3=7