IIT JEE , AIEEE Forum - Mathematics , Algebra

savvej

consider an equation: 2x² +mx +m² - 5 =0

(1)the set of values of m for which both the roots of the equation are less than 1 is:

(2)the set of values of m for which both the roots exceed -1 is:

savvej

plz give all the steps(rates gauranteed)

yash_pandit

Is the ans to (1) (-4,-2)

and(2) (4,infinty)

savvej

@yash your ans is not correct.

winner07

plz check dese,
if correct i will give solutions also

(1) ( -

, (-1-

13)/2 )

( (-1+

13)/2,

)

(2) ( (1-

29)/2 , (1+

29)/2 )

and plz give ratings if useful...

harsha_27

for real roots first of all D> or equal to 0
=>m

[-

(40/7),+

(40/7)]

1.for both roots of eqn. to be less than 1,
f(1)>0 and 1>-b/2a
=>1>-m/4
=>m>- 4
f(1)>0=>m²+m-3>0
=>m doesn't belong to (-1-

13/2,-1+

13/2)
=> m

(-

(40/7),-1-

13/2)

(-1+

13/2,

(40/7))
2.here f(-1)>0 and -1< -b/2a
=>-1<-m/4
=>m<4
f(-1)>0 =>m²-m-3>0
=> m

(1-

13/2,1+

13/2)
combining all these,m

(-

(40/7),1-

13/2)

(1+

13/2,

(40/7))

rate me iff satisfied......

winner07

oh yes i forgot to take dat

40/ 7 range
well other one is ok

harsha plz check ur other solution

savvej

wht if i tell u that both roots are less than 2,will u take f(2)> 0 and 2 <-b/2a ??
or is it valid only in this case????

harsha_27

yes .....of course , if u have written that roots are less than 2 then i would have taken that f(2)>0 and 2>-b/2a to solve the eqn. actually it doesn't depend on whether it is 2 or 1 it is true for any value given in that place......
here the coefficient of x2 is 2(means +ve) ....that means parabola opens upwards ......and in the qn. we want both roots < 1 (or any constant k)means 1 lies before both the roots on x-axis and let the roots be p and q (let p<q) ......f(x)>0 for x<p and <0 for p<x<q and >0 for x>q ,as we want both roots to be less than 1 =>1>q(>p) means f(1)>0 .....but if u consider only this condition ,it means u r considering only that at 1 ,fn. has a +ve value .....but this can happen in the interval x<p also which also gives a soln. for 1 less than both the roots also....so for 1 to be greater than both roots only,we'll take one more condition which removes that unwanted part.....that is 1>-b/2a ......now what is -b/2a?.....-b/2a is the point at which the fn. has an extremum (minima in this case as a is +ve) now if 1 is > this value also then obviously the soln. we'll get gives the values for which 1 is only >both roots thus removing the error and giving values what we want......so applying these 2 conditions we can solve the qn........hope u got it ...

harsha_27

i mean if we consider only f(1)>0 then we will be also considering the region when 1 is less than both the roots also in addition to the region when 1 is greater than both roots.....got it? because that region also satisfies this condition ......so in order that 1 to be greater than both roots we 'll take the condition that 1>-b/2a so that when we superimpose these two results we'll get only the region where 1 is > both the roots

savvej

can someone else try to explain???

harsha_27

you tell me where u have not understood?

atleast rate my efforts yaar

spideyunlimited

harsha is right :)..
wht didnt u get yaar

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