Saved Bookmarks
| 1. |
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10 in quadratic equations |
|
Answer» Let two parts be x and y.thereforex+y=20 ⇒ y=20-x Then we have 3x2-(20-x)=10 3x2+x-30=0 3x2+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. |
|