Question

let P and Q be the rots of quadratic equation

Answer 1

\[X ^{2} -(\alpha -2)X-\alpha-1=0. what is the minium possible value of p ^{2} + q ^{2}\]

Answer 2

X2−(α−2)X−α−1=0.what is the miniumpossible value of p2+q2

Answer 3

choices are
a.0
b.3
c.4
d.5

Answer 4

if p n q are the roots of the eq, then we have the eq
x^2-sum(p+q)x+product(pq)=0

Answer 5

n if the eqn is AX^2+BX+C=0 then
p+q=-B/A
pq=C/A

Answer 6

yaeh..but am not able to solve plz break down the steps

Answer 7

for the given eq
\[p+q=\alpha -2, pq=-\alpha-1\]

Answer 8

\[p^2+q^2=(p+q)^2-2pq\]

Answer 9

plug in the values n simplify

Answer 10

Hi,the value of pq will be 1-\[\alpha\]

Answer 11

choices are
a.0
b.3
c.4
d.5

Answer 12

how come -\[\alpha\]-1

Answer 13

it should be \[-1-\alpha\]

Answer 14

C=-alpha -1, A=1
product =C/A

Answer 15

thanks as its not in bracket...!

Answer 16

so it can't be \[-(\alpha-1)\]

Answer 17

\[p^2+q^2=(\alpha -2)^2-2(-\alpha-1)=\alpha^2-4\alpha=4+2\alpha+2\]

Answer 18

oh there is + in place of =

Answer 19

\[=\alpha^2-2\alpha+6\]

Answer 20

ohh gr8 thanks .. u deserve a Medel :)

Answer 21

but what about the choices u are given?

Answer 22

its 5

Answer 23

for minium value of alpha=1
minium function value will be 5 ...

Answer 24

exactly :), thats what i wanna ask, u deserve two medals, but only one can be awarded :)

Answer 25

α2−2α+6\[ so, (\alpha - 1)^{2} +5 => at \]

Answer 26

at alpha = 1 value of equation = 5 :)

Answer 27

right man :)