IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj_agarwal_1990

The no. of solutions of (3x²+6x+7)^1/2 + (5x²+10x+14)^1/2=4-2x-x² is?

[ans:2]

please post the solution at the earliest...

johri_anshuman

The graph indicates that there is only one solution.

I might be wrong.

saarika

the problem looks so complicated so it must be easy.u must be able to solve it directly since we are unable to solve it,it might not be having complicated answers.so try substituting some values.take the first part 3x²+6x+7. it is a perfect square for -1&5.substitute these values and see. they satisfy the eqn.and johri ur graph might be wrong.the ansis -1&5.

amaron

See,for such questions ,see if some expression is common in all the terms,for example,consider

(3x²+6x+7)^1/2 + (5x²+10x+14)^1/2=4-2x-x²

if you see,x² +2x is repeated therefore let it be y

implies

(3y +7)^1/2 + (5y+14)^1/2 = 4-y

squaring, you get

8y +21 +2((5y+14)^1/2(3y +7)^1/2)=16-8y+y²

squaring again,

(y² - 16y-5)² = 4(15y² +87y +98)

solving this is a tedious process,you will get 4 values of y of which you have to

take the real values only

and then equate y=x² +2x

again you will get corresponding values of x hence you can find the answer!

That solves the problem!

amar.gupta

Dear,

(3x²+6x+7)^1/2 + (5x²+10x+14)^1/2=4-2x-x²

(3x²+6x+7)^1/2 + (5x²+10x+14)^1/2=5-(1+x)²

(3x²+6x+7)^1/2 + (5x²+10x+14)^1/2+(1+x)²=5

if you give it a close look you find that (3x²+6x+7) and (5x²+10x+14) both are always +ve for every value of x as Descrimanant is -ve and coeff. of x^2 is +ve.

and min. value of both will be at x=-1 {check d/dx=0 at x=-1

and min.value of 3x²+6x+7 will be 4

and min. value of 5x²+10x+14 will be 9

so min. value of LHS will be 2+3+(1-1)²=5 at x=-1

so at x=-1 LHS=RHS

so it is a solution.

but otherwise for all other x, LHS >5

there will be only one solution possible.

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