Submitted by Edogowa Conan on Thu, 01/05/2017 - 11:26
Forums:
4x^2 +16xy +y^2 +24x =0. Prove that x is less than or equal to zero and greater than or equal to 2/5, and find similar restrictions for y.
I started by rearranging the above equation to (2x+y)^2 +12x(y+2), and then
(2x+y)^2 =-12x(y+2), so since LHS is positive, then RHS must be positive, but that can only happen if x is less than or equal to zero, (to cancel with the negative 12), or if (y+2) is less than or equal to zero, or y is less than or equal to-2. So I've proved one of the inequalities given. But how do I prove the other one.
Which STEP question?
Can you tell us which STEP question this is?
86 s1
86 s1
Spec paper
Spec paper
Whist your initial approach
Whist your initial approach is along the right lines (with completing the square), the way in which you do it makes life rather difficult for you at the end. In general, if possible, it's best to avoid mixing up your $x$'s and $y$'s together, you'd much rather have them separate, especially when trying to form inequalities with them.
To that end, you'd be best off completing the square as $(y+8x)^2 = 60x^2 - 24x$, then as you said, squares are always non-negative, so you get an easy quadratic inequality for $x$ that you can solve easily, giving you the requisite bounds.
With that in mind, I'll let you now (re-)complete the square such a way that you get $[g(x,y)]^2 = h(y)$ so that you can solve the (hopefully quadratic) inequality $h(y) \geq 0$.
Coefficient of x^n.
The next part of the question asks you to find the coefficient of x^n of the equation 9/[(2-x)^2 (1+x)]. This equation will become
9[(2-x)^-2 (1+x)^-1]. Expanding the first bracket, the coefficient of the x^n term is {[(-2)(-3)...(2)(1)]/n!} *([1/2]x)^n, and for the second bracket it is {[(-1)(-2)...(2)(1)]/n!} * (x)^n. What on earth is the answer to the multiple of those two numbers?
I'm not entirely sure what
I'm not entirely sure what you're trying to do, you might want to re-check the accuracy of those expansions. Also, if you're finding two terms of degree $n$ in both and multiplying them together, you'd get $2n$, not $n$. Instead, you want to pair off each term of order $k$ with a term of order $n-k$ in the other expansion.
86 s1 Specimen Paper
86 s1 Specimen Paper