SOLUTION: In one root ax^2+bX+c=0 is double the other.,prove that 2b^2=9ac ??????

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Question 818245: In one root ax^2+bX+c=0 is double the other.,prove that 2b^2=9ac ??????
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The solutions to ax%5E2%2Bbx%2Bc=0+ are x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ,
which I will temporarily abbreviate as x+=+%28-b+%2B-+sqrt%28DELTA%29%29%2F%282%2Aa%29+
with DELTA=b%5E2-4%2Aa%2Ac .

If %28-b%2Bsqrt%28DELTA%29%29%2F%282%2Aa%29=+2%28%28-b-sqrt%28DELTA%29%29%2F%282%2Aa%29%29 ,
-b%2Bsqrt%28DELTA%29=+2%28-b-sqrt%28DELTA%29%29
-b%2Bsqrt%28DELTA%29=+-2b-2sqrt%28DELTA%29
3sqrt%28DELTA%29=+-b which squares into 9DELTA=b%5E2 .

If %28-b-sqrt%28DELTA%29%29%2F%282%2Aa%29=+2%28%28-b%2Bsqrt%28DELTA%29%29%2F%282%2Aa%29%29 ,
-b-sqrt%28DELTA%29=+2%28-b%2Bsqrt%28DELTA%29%29
-b-sqrt%28DELTA%29=+-2b%2B2sqrt%28DELTA%29
-b=3sqrt%28DELTA%29 which squares into b%5E2=9DELTA .

Either way b%5E2=9DELTA which means b%5E2=9%28b%5E2-4%2Aa%2Ac%29 .
So b%5E2=9b%5E2-36%2Aa%2Ac%29 , and adding 36%2Aa%2Ac-b%5E2 to both sides we get
36%2Aa%2Ac=8b%5E2%29 , which dividing both sides by 4 gives us
9ac=2b%5E2%29 or highlight%282b%5E2=9ac%29 .