lf one root of quadratic equation ax²+ bx+c=0 is triple the other show that 3b²=1 ac
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-b+√(b^2-4ac)=3[-b-√(b^2-4ac)]
2b+4√(b^2-4ac)=0
b^2-4ac=(-b/2)^2
4b^2-16ac=b^2
3b^2=16ac
now !if ur question is incorrect?
2b+4√(b^2-4ac)=0
b^2-4ac=(-b/2)^2
4b^2-16ac=b^2
3b^2=16ac
now !if ur question is incorrect?
SanyamTaneja:
if not so, u can try multiplying 3 on other side
yes it's 3b²=16 ac
oh.. then thats ok
Answered by
0
b2 -4ac the solve and answer
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