IIT JEE , AIEEE Forum - Mathematics , Algebra

nishant13940r1

IF ONE ROOT OF THE EQUATION x²+px+1=0 IS 4, WHILE THE EQUATION

x²+px+q=0 HAS EQUAL ROOTS,THEN THE VALUE OF Q IS:

%SUGGESTED VALUES%

(A)49/4, (B)4/49, (C) 4, (D)NONE OF THESE

--THANKS AND RATES ASSURED

nivedh_89

let the other root of the 1st equation be a.......hence,

4+a=-p and
4a=1
therefore,
a=1/4 and -p=17/4

now for the 2nd equation,
-p=2b(where b is 1 root)
b=17/8
q=b²= (17/8)^{2

HENCE ANSWER IS D}

sboosy

substituting 4 in place of x we get p to b -17/4

now in the second quadratic since equal roots, discriminant =0

that is b ^2 = 4ac

so q= 289/64

nishant13940

not friend in arihant(praful kumar agarwal_)it is given 49/4

elessar_iitkgp

The solutions given by nivedh and sboosy are correct. Good work.

It seems that answer given in Arihant is incorrect

sboosy

Ya thats what i think.

sometimes even arihant can be wrong

nishant13940

ofcourse i also have checked thm

sboosy

Anyway i think there was no flaw in the method presented

fwks_phoenix · even i'm getting **q=289/64**

the method is correct

even i'm getting q=289/64 the ans is none of these

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