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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?

Jan 15, 2018

#1
+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\$.

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   Jan 15, 2018
#2
+1

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   Jan 15, 2018