Question

Let x be the smaller one of two consecutive integers. If the sum of the squares of the two integers is less than three times the product of the two integers by 1, then
A. x^(2)+(x+1)^(2)=3x(x+1)-1
B. x^(2)+(x+1)^(2)=3x(x+1)+1
C. 3[x^(2)+(x+1)^(2)]=x(x+1)-1
D. 3[x^(2)+(x+1)^(2)]=x(x+1)+1

Answer 1

x is smaller
so other = x+1

Answer 2

I know

Answer 3

where are u stuck ?

Answer 4

sum of the squares of the two integers = x^(2)+(x+1)^(2)

Answer 5

I know also

Answer 6

three times the product of the two integers = 3x(x+1)

Answer 7

sum of squares is less by 1
so
x^(2)+(x+1)^(2)=3x(x+1)-1

Answer 8

I am stuck in whether three times....

confusing.......wait a moment

Answer 9

three times the product of the two integers = 3x(x+1)

Answer 10

less than three times the product of the two integers by 1....
ok i got it.

Answer 11

so I got the answer A correctly ;D

Answer 12

good :)`

Answer 13

If 'x' is smaller of two consecutive integers. Then,
Larger integer=(x+1)

A/q,
=> (x)^2+(x+1)^2=3(x)(x+1)-1

Answer 14

Let x be the larger one of two consecutive odd numbers. If the sum of the squares of the two odd numbers is less than four times the product of the two odd numbers by 2, then

x^(2)+(x-2)^(2)=4x(x-2)-2
isn't it?

Answer 15

yes

Answer 16

yeah XD thank you~

Answer 17

you r correct @kryton1212 but i don't know that this is ur second question or first :/

Answer 18

sorry, thats correct

Answer 19

another question: Let k be a constant. Find the range of values of k such that the quadratic equation x^(2)+6x+k=3 has no real roots.
@jiteshmeghwal9 there are two questions and are separated :)
@hartnn never mind:)

Answer 20

for no real roots
D<0

Answer 21

D = b^2-4ac = ?

Answer 22

ohh ! okay @kryton1212 :)

Answer 23

@hartnn D=36-4k

Answer 24

36-4k <0

Answer 25

k>9

Answer 26

36-4k<0
-4k <-36
4k>36
yes k>9 is correct :)

Answer 27

you are so powerful .... thank you so much ;D

Answer 28

your welcome ^_^

Answer 29

and the following questions are about surds....

Answer 30

\[\frac{ 1 }{ 1+\sqrt{2} } +\frac{ 1 }{ \sqrt2+\sqrt{3} }+\frac{ 1 }{ \sqrt3+\sqrt{4} }+\frac{ 1 }{ \sqrt4+\sqrt{5} }=?\]

Answer 31

wow..

Answer 32

that will simplify greatly!

Answer 33

```
MATH PROCESSING ERROR
```

??

Answer 34

no its ok

Answer 35

But in my computer it's nt okay

Answer 36

refresh the page

Answer 37

I got the answer -1+sqrt5 but I don't know the way to do this...

Answer 38

if u don't know then how did u get ? :P

Answer 39

there are five choices in MC question. I calculate the question and answers:P

Answer 40

multiply and divide by conjugates