IIT JEE , AIEEE Forum - Mathematics , Algebra - AnS THIS SIMPLE QUES............ RATES FOR SURE!!!

krithika.r

(m+1)x²+ 2mx+5x +m+3 = 0 has equal roots............ find the value of m.

give a ans with solution............. rates for sure!!!!!!!!!!!!!!!!!!

krithika.r

i have made the correction

ganesha1991

equal roots D=0
D = b^2-4ac
a= (m+1)
b= 2m+5
c=m+3
substitute and find the answer
(2m+5)^2 = 4*(m+1)(m+3)
4m^2 +25+20m = 4m^2 + 12+16m
4m = -13
m=-13/4

pavann

ANS--- m=

rudra.panda

For equal roots the discriminant must be zero.

$b^2-4ac=0\\\\\Rightarrow b^2=4ac$

(m+1)x²+x( 2m+5) +(m+3) = 0

(2m+5)^2=4(m+1)(m+3)

$\Rightarrow 4m^2+25+20m=4m^2+16m+12$

$\Rightarrow 4m=-13$

$\Rightarrow m=\frac{-13}{4}$ (Answer)

rajatsen91

Re:AnS THIS SIMPLE QUES............ RATES FOR SURE!!!

(79)

the equation is [m+1]x^2+x[2m+5]+[m+3]
here a=m+1
b=2m+5
c=m+3
as the equation has real roots thr4
so b^2=[2m+5]^2=4m^2+25+20m
4ac=4[m+1][m+3]
b^2-4ac=0
so 4m^2+25+20m-4m^2-16m-12=0
4m+13=0
m=-13/4

IITPATIENT

If roots r = then; (sum of roots/2)^2 = product of roots; put value we get m=-13/4

(2)

equal roots D=0
D = b^2-4ac
a= (m+1)
b= 2m+5
c=m+3
substitute and find the answer
(2m+5)^2 = 4*(m+1)(m+3)
4m^2 +25+20m = 4m^2 + 12+16m
4m = -13
m=-13/4

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