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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?

Jdaye  Jan 15, 2018
 #1
avatar+93015 
+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

 

 

cool cool cool

CPhill  Jan 15, 2018
 #2
avatar+93015 
+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

 

 

cool cool cool

CPhill  Jan 15, 2018

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