+0

0
353
3
+81

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.

Jun 25, 2018
edited by HelpPls  Jun 25, 2018

#1
+102905
+3

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

Jun 25, 2018
#2
+102905
+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

Jun 25, 2018
edited by CPhill  Jun 25, 2018
#3
+81
+1

Ok, thanks so much!!!

HelpPls  Jul 3, 2018