SOLUTION: Find the quadratic equation whose roots are twice the roots of 3x²+19x-14=0

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Question 885977: Find the quadratic equation whose roots are
twice the roots of 3x²+19x-14=0
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B19x-14%22%22=%22%22%220%22

Let y%22%22=%22%222x

Then x%22%22=%22%22y%2F2

Substitute:

3%28y%2F2%29%5E2%2B19%28y%2F2%29-14%22%22=%22%22%220%22

3%28y%5E2%2F4%29%2B19y%2F2-14%22%22=%22%22%220%22

Multiply through by 4

Answer:

3y%5E2%2B38y-56%22%22=%22%22%22%220%22%22

-------------------

Checking:

Solve:

3x%5E2%2B19x-14%22%22=%22%22%220%22

%28x%2B7%29%283x-2%29%22%22=%22%22%220%22

x+7 = 0;  3x-2 = 0
  x = -7    3x = 2
             x = 2%2F3

Solve   

3y%5E2%2B38y-56%22%22=%22%22%22%220%22%22

%28y%2B14%29%283y-4%29%22%22=%22%22%220%22

y+14 = 0;   3y-4 = 0
   y = -14    3y = 4
               y = 4%2F3

The roots of the second one are twice the roots of the first.

Edwin