Question

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Answer 1

Let a, b, c be the sides of a triangle. No two of them are equal and λ ∈ R. If the roots of the equation x^2 + 2(a + b+ c) x
+ 3λ (ab + bc + ca) = 0 are real, then

(A)\[\lambda <\frac{ 4 }{ 3 }\]
(B)\[\lambda>\frac{ 5 }{ 3 }\]
(C)\[\lambda \epsilon \left( \frac{ 1 }{ 3 },\frac{ 5 }{ 3 } \right)\]
(D)\[\lambda \epsilon \left( \frac{ 4 }{ 3 },\frac{ 5 }{ 3 } \right)\]

Answer 2

@vishweshshrimali5

Answer 3

@mathslover

Answer 4

I have a start

Answer 5

Okay ?

Answer 6

If the roots are real then D should be greater than or equal to 0

Answer 7

yes

Answer 8

\[\huge 4(a+b+c)^{2}-12\lambda(ab+bc+ca) \ge 0 \]

Answer 9

ok

Answer 10

The next step?

Answer 11

Isolate lambda

Answer 12

probably

Answer 13

First of all divide by 4 on both sides of the inequality, you will get :

\(\huge (a+b+c)^{2}-3\lambda(ab+bc+ca) \ge 0\)

Answer 14

Now expand (a+b+c)^2

Answer 15

Yes ! i was about ti say it

Answer 16

Expand ok wait

Answer 17

\[\huge a ^{2}+b ^{2}+c ^{2}+2(ab+ac+bc)\]

Answer 18

Yes

Answer 19

After that

Answer 20

Separate similar terms

Answer 21

yes wait

Answer 22

\[\huge (2-3\lambda)(ab+bc+ac)+(a ^{2}+b ^{2}+c ^{2})\ge 0 \]

Answer 23

^ is it right

Answer 24

are you there

Answer 25

@vishweshshrimali5

Answer 26

Yeah sorry

Answer 27

Okay

Answer 28

Is it right?

Answer 29

Yes

Answer 30

Take (2 - 3lambda) part in RHS

Answer 31

a+b>c
a+c>b
b+c>a

Answer 32

(2−3λ)(ab+bc+ac) in RHS ?

Answer 33

@ikram002p that would help

Answer 34

@ikram002p but how are we going to use the triangle inequality.

Answer 35

ikr ;)

Answer 36

Wait that can be looked afterwards let so I put that in RHS

Answer 37

Don't use it now, you would make the present solution much more complicated.