IIT JEE , AIEEE Forum - Mathematics , Algebra

comeon

the roots of x² + bx + c = 0 are both real and greater than 1.

if s = 1+ b+ c,

then comment on the value of s

nivedh_89

s>0..........!!!!!!!!!
bcas b = a negative no.........

and always for the roots wch r real n greater than 1,1+c is always greater than b......!!!!!!!!
if the roots r less than 1,then their sum maybe greater than their product but they will b greater in a margin less than 1....so 1+product>sum.....and if the roots r greater than 2,their products r greater than their sums..............so s>0................!!!!!!!!!!!!

rajatsen91

Nivedh your giving too much effort in this problem.
if f(x)=x^2+bx+c then f(1) = 1+b+c
As both roots are greater than one then from the graph we get that s>0.

{note as the coefecient of x²is >0 so the graph opens upwards}

raghavnegi

ans is s>0

spideyunlimited

s>0 ... n yes gud thinking rajatsen but it can be done the other way too by seeing that c >1 and b<-2 and then checking and finding that s > 0 always

metal

I cant understand how you all are solving it.

we know thatb<-2, i.e. 1+b<-1 and c>1

so 1+b+c can be >0 or =0.

Consider this expression (x-1)(x-2)=x²-3x+2

In this b= -3 and c = 2.

So, 1+b+c = 1-3+2 = 0. [which contradicts your answers]

Note that, s cannot be < 0, becuse sum of two real numbers cannot be greater than their product by a number greater than 1

So, the answer is:-

S>0 or = 0

please rate me.

spideyunlimited

uf oh metal b<-2 and c >1
so 1+b+c > 0 . where the hell did u get the equality sign from mate... wht r u tryin to do..
it will only be 0 when b = -2 and c = 1 but that is not allowed here.. see the first line b<-2 and c >1

metal

Spidey dear,

haven't you read what I have written?????????

Who the hell told you that 1+b+c will be equal to 0 when b= -2 and c= 1.?????

Of course that is a case, but this is not valid here.

But there are other possible values of b and c such that b<-2 and c>1, yet 1+b+c=0.

I even gave an example, and I am giving it here again:-

Consider the expression x²-3x+2;

where b= -3 which is obviously less than -2, and c=2, which is also obviously greater than 1, yet 1+b+c=1-3+2=0.

GOT IT????????

[correct me if I am wrong]

If I am correct please rate me {I am currently in need of some good rates, pal. Please help}

neeraj_agarwal_1990

If both roots of ax^2 +bx +c=0 are greater than k...
then a.f(k)>0 is the necessary and sufficient condition...

so 1.(1+b+c)>0

so s>0

spideyunlimited

MR METAL!!! ... in the equation u have given x^2 - 3x + 2... what are the two roots?
x = 1, and x = 2
but the question clearly states that an equation where both roots are greater than one..ur equation has one root equal to one. so there!

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