+54 1.)
Let $d$ and $q$ be the roots of the quadratic equation $x^2+7x-4=0$. Find the value of $2d^2+19d+5q-8$.
2.)
How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following property: the sum of three things -- the hundreds digit, the units digit, and the product of the units and tens digits -- is eight more than the tens digit?
3.)
The terms of a particular sequence are determined according to the following rules:
* If the value of a given term is an odd positive integer , then the value of the following term is
* If the value of a given term is an even positive integer , then the value of the following term is .
Suppose that the terms of the sequence alternate between two positive integers . What is the sum of the two positive integers?
+129852 Let $d$ and $q$ be the roots of the quadratic equation $x^2+7x-4=0$. Find the value of $2d^2+19d+5q-8$.
x^2 + 7x - 4 = 0
The roots are
[-7 + √ [ 7^2 + 16 ] ] / 2 = [-7 + √ [ 65 ] ] / 2 = d
And
[-7 - √ [ 7^2 + 16 ] ] / 2 = [ - 7 - √65] / 2 = q
So......
2d^2 + 19d + 5q - 8 =
(1/2) [ 49 - 14√ 65 + 65] + (19 / 2) [-7 + √ [ 65 ] ] + (5/2) [ - 7 - √65] - 8 =
49/2 - 7√ 65 + 65/2 - 133/2 + (19/2)√ 65 - 35/2 - (5/2)√ 65 - 8 =
[ 49 + 65 - 133 - 35 -16] / 2
[ 114 - 184 ] / 2
[-70] / 2 =
-35
BTW......the results are the same no matter which root we assign to d, q
+129852 2 )
How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following property: the sum of three things -- the hundreds digit, the units digit, and the product of the units and tens digits -- is eight more than the tens digit?
Let a = 100's digit b = 10's digit c = units digit
And we have that
a + c + bc = b + 8
a + c + bc - b = 8
a + c + b (c - 1) = 8
Here are the possibilities
a b c
2 4 2
3 1 3
4 0 4
