IIT JEE , AIEEE Forum - Mathematics , Algebra

Utsav55

${x}^{2}$ - (a+b)x + ${a}^{2}$ - ab + ${b}^{2}$ is:

Options:

>=0

>0

<=0

<0

Plz explain.

THANKS

chinnu

if a is not equal to b then option 2 is correct

add&sub ab to the eqn

u get (x-a)(x-b)+(a-b)^2

min value of (x-a)(x-b) is -[a-b/2]^2

hence the total value is more than zero

plz rate if its correct

mahtwog

discriminent is < 0 and co-efficient of x² is >0

Hence it is >0 [Short and sweet]

dakshayadav

i think da answer is >0.

the value of discriminant is coming out to b = -3(a-b)(a-b)

and the coeff of x2 is positive. so the value will always b positive.

Utsav55

how can u say that D < 0
Plz explain this.

dakshayadav

the value of d is coming out to b -3(a-b)square.

ravi00

are if D<0 den d root will be negative so x<0
plz correct me if i m wrong.....

allamraju

The given expression is [x-(a+b/2)]²+(3/4)[a-b]² >0 for all x if a and b are unequal.Ravi,Hope u understood.

ashish040191

D<0 means x hv no real solution i.e graph of given eqn don't toch or cut x axis at any point.

coeff of x² is > 0 means the graph is in upward ditrcn. so it ll b always>0

Utsav55

thank you guys 4 the reply.

Prob solved.

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The given expression is [x-(a+b/2)]2+(3/4)[a-b]2 >0 for all x if a and b are unequal.Ravi,Hope u understood.

The given expression is [x-(a+b/2)]²+(3/4)[a-b]² >0 for all x if a and b are unequal.Ravi,Hope u understood.