AboutAbe Mantell Expertise Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
Question (Quadratic Equation)
1)Solve these equation.(use 2^x=y) 2^2x-9(2^x)+20=0
2)if p and q are the root for quadratic equation 2x²-5x-1=0,then find the value for
a)p²+q²
b)p over q + q over p
(Function)
3)The function f and g are defined by f:x=x²+3 and g:x=2x+1.Find the range of fg.
4)The function f and g are defined by f:x=3x-1 and g:x=x²+1.Find the range of g.
5)A function is defined by f(x)=x²-6,x>6. Find the valus of f(x) and f^-1(x) on the same axes.
Answer Wow, you sure have many questions! How about I do a couple and see if you can do the others?
1. Do you mean 2^(2x)-9(2^x)+20=0, if so then if y=2^x, we get
. y^2-9y+20=0 ==> (y-4)(y-5)=0 ==> y=4,5 ==> 2^x=4 or 2^x=5
. ==> x=2 or x=ln(5)/ln(2)
2. Divide 2x²-5x-1=0 by 2 to get x²-(5/2)x-1/2=0...if p & q solve this,
. then it factors to (x-p)(x-q)=x^2-(p+q)x+pq...so, p+q=5/2 and pq=-1/2
. If p+q=5/2, then (p+q)^2=25/4 ==> p^2+2pq+q^2=25/4...but since pq=-1/2
. 2pq=-1...so p^2-1+q^2=25/4 ==> p^2+q^2=29/4
. p/q + q/p = (p^2+q^2)/(pq) = (29/4)/(-1/2) = -29/2
