Asked by Nana ama
If one root of ax+bx+c=0 is treble the other prove that 3b²-16ac=0
Answers
Answered by
oobleck
since the roots are (-b±√(b^2-4ac)/2a, you need
-b+√(b^2-4ac) = 3(-b-√(b^2-4ac))
2b = -4√(b^2-4ac)
4b^2 = 16b^2 - 64ac
12b^2 - 64ac = 0
3b^2 - 16ac = 0
-b+√(b^2-4ac) = 3(-b-√(b^2-4ac))
2b = -4√(b^2-4ac)
4b^2 = 16b^2 - 64ac
12b^2 - 64ac = 0
3b^2 - 16ac = 0
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