SOLUTION: Divide 20 into two parts such that three times the square of one part exceeds the other part by 10

SOLUTION: Divide 20 into two parts such that three times the square of one part exceeds the other part by 10

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Question 964028: Divide 20 into two parts such that three times the square of one part exceeds the other part by 10
Answer by amarjeeth123(569) (Show Source): You can put this solution on YOUR website!
Let the two parts be x and (20-x).
Then we have 3x^2-(20-x)=10
3x^2+x-30=0
3x^2+10x-9x-30=0
x(3x+10)-3(3x+10)=0
(x-3)(3x+10)=0
x=3 is the solution.
The two parts are 3 and 17 respectively.
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