IIT JEE , AIEEE Forum - Mathematics , Algebra

shank

If p,q,r are real and p

q,then show that the roots of the equation

(p-q)x²+5(p+q)x-2(p-q)=0

are real and uneqal.

aditya_arora04

Hi,

Look,

Calculate the roots by formula : [-b

(D)^1/2] / 2a

So, Now, D = 33(p² + q²) + 34pq

which is real .

so, its root is also gonna be real. Now, we can well see that the roots are different.

Arjun

hiiiiiiiii..........

hers it........

divide throughout by p-q

x²+5(p+q)/(p-q)x-2=0

now use da determinant condition.........

D=^{[ ]}

b²-4ac=^{[ ]}

[(p+q)/(p-q)]²- 4*-2

D=

[(p+q)/(p-q)]²+8

under root everythin is [(p+q)/(p-q)]²>0 as both are squared nd dis is added to a positive number 8

so under root everythin is positive.............

so therefore D>0

since D>0 the roots are are real..........nd since p

q,the roots are unequal............

hope u got it......

thank u.... :)

akku

Hi Shank

DISCRMINANT=b^2-4ac=25(p+q)^2+8(p-q)^2

=33(p^2+q^2)+34>0 thus the roots are real and unequal (asDnot=0 )

(D=0 is the condition for equal roots)

ALTERNATIVELY for proving roots unequal

let roots be x1,x2

x1+x2=-5(p+q)/(p-q)

x1x2=-2

(x1-x2)^2=(x1+x2)^2-4x1x2=25(p+q)^2+8(p-q)^2/(p-q)^2

which is non zero HENCE ROOTS ARE UNEQUAL

amar.gupta

good work akku,