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Mathematics 17 Online
OpenStudy (sugarmochi):

Please help, its urgent :((( The sum of the roots of x^2 +px+2=0 is equal to twice the product of the roots of 3x^2 +4x-2(p+1) =0. Find the values of P

OpenStudy (welshfella):

Sum of roots = -b/a Product = c/a Where a b and c are the coefficients in standard quadratic equation

OpenStudy (sugarmochi):

so would it be -p/1 = 2(-2/3) ?

OpenStudy (welshfella):

The -p/1 is correct But c = -2(p+1)

OpenStudy (welshfella):

And the a in second equation is 3

OpenStudy (sugarmochi):

how would i solve this though- can i do it straight as -p = 2(p+1) ?

OpenStudy (sshayer):

\[Product~of~roots=\frac{ c }{ a }\]\[Given~eq.~is~3x^2+4x-2(p+1)=0\] \[Product ~of~roots=\frac{ -2(p+1) }{ 3 }\] \[\frac{ p }{ 1 }=2*\frac{ -2(p+1) }{ 3 }\] 3p=-4p-4 find p

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