IIT JEE , AIEEE Forum - Mathematics , Algebra

littlebrother91

the value of (a) for which the sum of the squares of the roots of the quadratic equation x2 -(a-2)-a-1=0 assumes the least value is

a) 0

b)1

c)2

d)3

anchitsaini

I am assuming the quadratic is

Then

$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2 \\ \\=(a-2)^2+2(a+1) \\ \\=a^2-2a+6$

which is minimum when

$\frac{d}{da}of \ the \ above \ equation \ is \ 0 \\ \\cos \ the \ second \ differential \ of \ the \ equation \ is \ positive$

or

$2a-2=0 \\ \\a=1$

deedee

let A and B be d roots

thus A+B=a-2

AB=-a-1

A^2+B^2=(A+B)^2-2AB

solvin u get

A^2+B^2=a^2-2a+6=f(x)

this shud hav d least value thus f'(x)=0

thus

2a-2=0 thus a=1

@anchitsaini sorry ...ni dekha ki u posted d sol

akhil_o

$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\\=b^2/a^2-2c/a$

${(a-2)}^{2}+2(a+1)\\=a^2-2a+6\\\\differentiating\; for \;minima \\2a-2=0 \\a=1$

akhil_o

oops sorry.....
wasnt used to new editor...
didnt see above posts
sorry again

oneyeartogo

an alternate method:

Same steps as above till we get

a²-2a+6

= [a²- 2a + 1] - 1 + 6

= [a-1]² + 5

which will have minimum value if [a-1]² is smallest i.e. a = 1 and [a - 1]² is 0

So a = 1

Conjurer

Just to pass time, another method:

Let a^2 - 2a + 6 =y

=> a^2 - 2a +6-y =0

D>=0

=> 1-(6-y)>=0

y>=5