Question

Given that a/3 and b/3 are the roots of the quadratic equation mx(2x-3)=10x-n,calculate the values of m and n if a+b=19/2 and ab=45/2.
@rational

Answer 1

\[2mx^2-3mx-10x+n=0\]\[2mx^2-(3m+10)x+n=0\]

Answer 2

Yes use the vieta's formulas

sum of roots :
\[\frac{a}{3}+\frac{b}{3}=3m+10\]

product of roots:
\[\frac{a}{3}\cdot\frac{b}{3}=n\]

Answer 3

\[\frac{ a+b }{ 3 }=3m+10\]\[\frac{ ab }{ 9 }=n\]

Answer 4

sholud it be \[SOR=-\frac{ b }{ a }\]\[POR=\frac{ c }{ a }\]?
@rational

Answer 5

Oops yes, my mistake!

Answer 6

thnks for catching :)

Answer 7

So after fixing that mistake, do we get

sum of roots :
\[\frac{a}{3}+\frac{b}{3}=\frac{3m+10}{2m}\]

product of roots:
\[\frac{a}{3}\cdot\frac{b}{3}=\frac{n}{2m}\]

Answer 8

\[\frac{ a+b }{ 3 }=\frac{ 3m+10 }{ 2m }\]\[\frac{ ab }{ 9 }=\frac{ n }{ 2m }\]

Answer 9

should we substitute a+b=19/2 and ab=45/2 into the equations? @rational

Answer 10

Yes

Answer 11

i got \[m=-\frac{ 15 }{ 4 } and~n=-\frac{ 75 }{ 4 }\]but the answer from the book say it is \[m=3,n=15\]

Answer 12

Check your work, you will get 20m = 60 or m=3

Answer 13

\[-\frac{ 3m+10 }{ 2m }=\frac{ 19 }{ 6 }\]\[-18m-60=38m\]\[m=-\frac{ 15 }{ 4 }\]