+0

0
104
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Let x^2+bx+c=0 be a quadratic whose roots are each two more than the roots of 3x^2-5x-7. What is c?

Dec 16, 2020

#1
+117576
+1

By Vieta.....in the last quadratic.....call the roots  m ,n

Sum of the  roots  =  m + n =  5/3

Product  of the  roots  = mn = -7/3

Product of  the roots  = c

(m+ 2) ( n + 2)   =  c

mn + 2m + 2n+ 4    =  c

-7/3 + 2 ( m + n)  + 4  = c

-7/3 + 2 ( 5/3)  + 4   =c

-7/3  + 10/3 + 4   = c

1 + 4   =  c

5    = c

CORRECTED   !!!

Dec 16, 2020
edited by CPhill  Dec 16, 2020
#2
0

Oh my gosh, CPhill! I'm a huge fan! I hate to say this, but that answer was wrong, the correct answer is 5. When I put it in it said (-7/3)+2(5/3)+4=5.

Guest Dec 16, 2020
#3
+117576
+1

Duh!!!!   on me.......just a sign error....

Should be

(-7/3) + 10/3  + 4  =

1  +  4   =

5

Now we agree  !!!....thx for the  correction  !!!

CPhill  Dec 16, 2020
#4
+1

Wow, well CPhill even if its wrong that's still really cool that you know how to do that!

Dec 16, 2020
#5
+117576
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I'm pretty sure that it's correct....LOL!!!!

CPhill  Dec 16, 2020
#6
+1

The roots of the second equation are  5/6 +- sqrt (109)/6    via Quadratic Formula

add 2 to each of these roots to get

17/6 +- sqrt(109) /6

multiply them together to get c    289/36 - 109/36 = 180/36 = 5

Dec 16, 2020
#7
0

Dec 16, 2020
#8
+117576
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No man....you pointed my stupid  "addition" error out to me.....the thx is to  you  !!!!

CPhill  Dec 16, 2020