IIT JEE , AIEEE Forum - Mathematics , Algebra - Progression-2

ramyadiamond

If a,b,c,d are rational and are in GP then the roots of the equation (a-c)²x² +(b-c)²x +(b-d⁾² - (a-d)²=0 are necessarily

a) imaginary

b)irrational

c)rational

d) real and equal

ramyadiamond

experts plzz help

ramyadiamond

somebody plzz try this out. I want the answer soon. so plz plz helpp

kiran

Hi Ramya

The question is easy but given in thast way to confuse.

Let a general quadratic equation be

ax²+bx+c =0

The equation has Real roots if b² - 4*a*c >= 0 .

By closely Analysing the above condition we can conclude that for a equation to have real roots the necesarry conditions are :

1) If 'a' and 'c' both are +ve , then b² > 4*a*c

2) Any one of ' a ' or ' c ' should be negative . ( Then b² +4*a*c > 0 always)

So in the question given observe that ' a ' is positive always . 'a' = (a-c)² >0 .

So If the roots are to be real then either ' c ' should be -ve or b² > 4*a*c

' c ' = (b-d⁾² - (a-d)²= ( b - d + a - d) ( b - d - a + d)

= ( b + a - d ) ( b - a)

Let the GP be a , ar , ar² , ar³ where r is the common ratio.

( b ,c, d are ar , ar² , ar³ )

' c ' = ( ar + a - ar³ ) ( ar - a)

= a²(r + 1 - r³ ) ( r - 1)

Now When r > 1

(r-1) >0 but (r + 1 - r³ ) < 0 ( check urself!)

Hence ' c ' < 0

When r < 1

(r - 1) < 0 and (r + 1 - r³ ) > 0 ( Check urself! )

Hence ' c ' < 0

When r = 1 then

' c ' = 0 .

So in all cases of r

b² > 4*a*c

which implies b² - 4*a*c > 0

And Hence the roots are rational. c is the answer .

Please tell whether I am right .

Cheers!

ramyadiamond

thanks so much, its absolutely right.