+0  
 
0
85
3
avatar+33 

Please help me with this question: Thanks!

 

Problem 1:

Problem 1

I tried but I got the wrong answer. (I got 2 1/2)

 

2:

 

 

Thanks so much!😀

Please explain how to do them so I know how to do them next time.

HelpPls  Jun 25, 2018
edited by HelpPls  Jun 25, 2018
 #1
avatar+88962 
+2

This isn't as difficult as  it seems...we just need to use some "creative" Algebra

 

4x^2 +9x - 6 =  0

 

In the form  ax^2  + bc  + c  = 0

 

The sum of the roots,  ( s + t )  =   -b/ a =    -9 / 4

The product of the roots,   (st )  = c /a = -6/4  = -3/2

 

Note that  we can write

 

s/t  +  t /s     =   [ s^2  + t^2 ] / st

 

And note that  ( s + t)^2  = s^2  + 2st + t^2

So

(-/9/4)^2  = s^2  + 2(-3/2) + t^2

 

81/16  =  -3  +  [s^2  + t^2]  .....    add 3 to both sides

 

81/16 + 3  = s^2 + t^2

81/16 + 48/16 = s^2 + t^2

129/16  = s^2 + t^2

 

So......

 

s/t + t/s  =   [ s^2 + t^2 ]/ st   =  [ 129/16] / [-3/2)]  =  [129/16] * [ -2/3]  = -43 / 8

 

cool cool cool

CPhill  Jun 25, 2018
 #2
avatar+88962 
+3

Here's the second one

 

2x^2 - 8x  +  7  =   0

 

Let's write this in the form  mx^2 + nx  + q  = 0    to avoid confusion with the roots

 

And we can write 

 

1 / (2a)   +  1 / (2b)  =    [a + b ] / [ 2ab ]

 

Using a similar idea as in the first one

 

The sum of the roots ( a + b)  = -n/m =  - (-8) / 2  = 8/2  = 4

The product of the roots  ab  = q/m    = 7/2

 

So...

 

1/(2a) + 1/ (2b)   =[ a + b ] / [2ab]  =  4 / [ 2 (7/2) ] =  4 / 7 

 

 

 

cool cool cool

CPhill  Jun 25, 2018
edited by CPhill  Jun 25, 2018
 #3
avatar+33 
+1

Ok, thanks so much!!!

HelpPls  Jul 3, 2018

41 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.