Question

if one root of ax2 +bx+c =0 is double of other then show 2b^2=9ac​

Answer 1

Answer:

HENCE PROVED IN SOLUTION..

Step-by-step explanation:

*********** HEY DEAR *************

HERE IS YOUR ANSWER...

LET ONE ROOT = X

OTHER ROOT = 2X

SUM OF ROOTS = -B/A =>  3X

                    3X = -B/A

                    X = -B/3A

                 ( X^2) ={ (B^2) /9(A^2)} ..................................(1.)

PRODUCT OF ROOTS = C/A

                   2(X^2) = C/A

                   (X^2) = C/2A ........................................(2.)

EQUATE 1. AND 2.

           

                    (B^2)  /  9(A^2) = C/2A.

BY INTERCHANGING THE SIDES AND SOLVING..

                      2 (B^2) = 9AC

.............................HENCE PROVED...........................

HOPE IT HELPS YOU..

THANKS..

^_^

Answer 2

Let m and 2m be the roots

Their sum is m+2m=-b/a
3m=-b/a

m= -b/3a

Product= c/a
mx2m=c/a
2m^2 =c/a

2(-b/3a)^2 = c/a
2b^2 / 9a^2 = c/a
2b^2 = 9ac
Hence proved