Quadratic inequalities stumped!
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Dr.Peterson
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Where are you stuck? Please show your work so far, so we can help you over the next step, or find an error in what you've done.
pka
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It is not clear, to me at least, the restrictions. So let's agree that \(\displaystyle a,~m,~x,~\&~y\) are all positive.
Moreover, we are given:
\(\displaystyle \begin{align*}1.~x+y&=a \\2.~\frac{1}{x}+\frac{1}{y}&=1\\3.~\frac{y}{x}&=m \end{align*}\)
To gaz5563: these questions are for you to answer.
Using 1. & 2. Prove that \(\displaystyle xy=a\).
From 3. does it follow that \(\displaystyle y=mx\) or \(\displaystyle a=mx^2~?\)
Does that mean that \(\displaystyle x=\frac{\sqrt a}{\sqrt m}~?\)
What you can do with all of that. i.e. what is \(\displaystyle y=~?\)
Dr.Peterson
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It certainly isn't obvious what method they expect you to use to "eliminate x and y"; different methods will (initially) result in what appear to be very different equations.
One nice approach uses the given y/x = m and pka's xy = a. If you solve each of these for x and set those expressions equal, you can obtain an expression for y in terms of a and m; then do the same with y to solve for x. Then put these into one of the given equations, and you'll have an equation in only a and m. This may take several forms; they can be simplified, but any of them should be acceptable.
It isn't clear what you are to assume, as pka pointed out; they don't seem to want to assume even that a is real until the last question, but I found myself wanting to assume at least that x and y are positive real numbers. Of course, they certainly can't be zero. Did you copy the entire problem exactly as given?
One nice approach uses the given y/x = m and pka's xy = a. If you solve each of these for x and set those expressions equal, you can obtain an expression for y in terms of a and m; then do the same with y to solve for x. Then put these into one of the given equations, and you'll have an equation in only a and m. This may take several forms; they can be simplified, but any of them should be acceptable.
It isn't clear what you are to assume, as pka pointed out; they don't seem to want to assume even that a is real until the last question, but I found myself wanting to assume at least that x and y are positive real numbers. Of course, they certainly can't be zero. Did you copy the entire problem exactly as given?
The book answers are (m+1)^2 =ma and a less than or equal to 0 and a greater than or equal to 4 cant quite figure what they have done ,the only thing missing is where in front of a does not equal 0 the book is bostock Chandler pure mathematics 1 ex 3c no 16
Dr.Peterson
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That's the equation I get. What exactly are you asking? As I understand it, you are saying that the only thing you omitted from the problem is a given condition that a≠0; and that the answer to the second part is "a≤0 or a≥4". I haven't yet looked into that part; but it seems odd that the answer would include the forbidden a=0!
Did you try the ideas we've offered? Please show your work, so we can see where you need help.
Did you try the ideas we've offered? Please show your work, so we can see where you need help.