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Mathematics 13 Online
OpenStudy (anonymous):

Find roots if two roots are equal : 4x^3+20x^2-23x+6=0

OpenStudy (anonymous):

pls use the relation: Roots: p,q,r p+q+r= -b/a pq+qr+pr= c/a pqr = d/a

OpenStudy (anonymous):

If two roots are equal then: \[2p + r = -5\] \[\large p^2r = \frac{3}{2}\] Using these two equation find p and r.. Can you do it??

OpenStudy (anonymous):

no

OpenStudy (anonymous):

full sol. pls

OpenStudy (anonymous):

as p=q, p+q=2p nw 2p +r=-5------1) nd p^2r=3/2------2) frm 1) r=-5-2p-----3) substitute 3)in 1)

OpenStudy (anonymous):

den u'll get p^2(-5-2p)

OpenStudy (anonymous):

we cant substitute 3) in 1)

OpenStudy (anonymous):

=3/2

OpenStudy (anonymous):

ya itz upto u

OpenStudy (anonymous):

wait wait i dnt think u can do dat

OpenStudy (anonymous):

lyk u get a value frm one equation nd hw can u substitute it in de same thing?

OpenStudy (anonymous):

yes thats what i was saying

OpenStudy (anonymous):

i mean 3) in 2)

OpenStudy (anonymous):

well i got de equation no: wrong bt wat i did was ryt

OpenStudy (anonymous):

den u can solve de quadraitic equation

OpenStudy (anonymous):

the full sol. !

OpenStudy (anonymous):

can u tell me de value of p? lyk i'm already workin on sme problemz

OpenStudy (anonymous):

you mean the ans.?

OpenStudy (anonymous):

solve fr p...p^2(-5-2p)=3/2

OpenStudy (anonymous):

p is root we have to find p q and r..

OpenStudy (anonymous):

if v get p, den (x-(p)) is a factor of the givn polynomial den v can divide de polynomila by (x-(p)) to get a quadratic polynimial den by splitting de middle term v get q nd r

OpenStudy (anonymous):

i don't want the ans. from synthetic division.

OpenStudy (anonymous):

bt v hav used de co efficient- zero relation ryt nd den only v r dividing...well dis is all i hav learnt

OpenStudy (anonymous):

i guess u hav to bump de question to get de ans: srry

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

@ganeshie8 @mukushla @nitz Does any one of you knows Theory of Equations?

OpenStudy (jiteshmeghwal9):

do u mean proof of quadratic formula???

OpenStudy (anonymous):

no

OpenStudy (jiteshmeghwal9):

then wht???

OpenStudy (anonymous):

google it.

OpenStudy (jiteshmeghwal9):

http://en.wikipedia.org/wiki/Quadratic_equation

OpenStudy (anonymous):

ok see this Let's Suppose p=q form p+q+r= -b/a pq+qr+pr= c/a u have 2p+r=-5 -------->4p+2r=-10 -------->p+2r=-10-3p (I) p^2+2pr=-23/4-------->p(p+2r)=-23/4 (II) plug equation (I) in equation (II) u will get p(10+3p)=23/4 its a simple quadratic equation and it gives p=1/2 and p=-23/6

OpenStudy (anonymous):

roots : 1/2, 1/2 , -6

OpenStudy (anonymous):

Yeah I got it...

OpenStudy (anonymous):

@shubham.bagrecha p=1/2 and r=-5-2p=-6

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

@mukushla why p=q=-23/6 and c=8/3 not taken??

OpenStudy (anonymous):

when p=1/2,c=-6 when p= -23/6, c=8/3 ??

OpenStudy (anonymous):

check it with p^2 r=3/2

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

got that

OpenStudy (anonymous):

K

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