neokeynesian

10-29-2004, 06:08 AM

Please see the attachment

For question 2,

(2.sqrt(ab))^2 = 2.ab not 4.ab??

anyone can explains?

thanks

For question 2,

(2.sqrt(ab))^2 = 2.ab not 4.ab??

anyone can explains?

thanks

View Full Version : pp question - square roots

neokeynesian

10-29-2004, 06:08 AM

Please see the attachment

For question 2,

(2.sqrt(ab))^2 = 2.ab not 4.ab??

anyone can explains?

thanks

For question 2,

(2.sqrt(ab))^2 = 2.ab not 4.ab??

anyone can explains?

thanks

lmtuan

10-29-2004, 07:06 AM

question 1

[1 / sqrt(5) ] * { [1 + sqrt(5)]/2 - [1 - sqrt(5)]/2 }

= [1 / sqrt(5)] * { [1 + sqrt(5) -1 + sqrt(5)] / 2}

= [1 / sqrt(5)] * { [2 * sqrt(5)] / 2}

= [1 / sqrt(5)] * sqrt(5)

= 1

Thus, answer is B

question 2:

Note that a>0 and b>0 , then sqrt(ab) = sqrt(a)*sqrt(b), sqrt(a)>0, sqrt(b)>0, and [sqrt(a) + sqrt(b)]>0.

We have:

sqrt{a + b + 2*sqrt(ab)}

= sqrt{ [sqrt(a)]^2 + 2*sqrt(a)*sqrt(b) + [sqrt(b)]^2 }

= sqrt{ [sqrt(a) + sqrt(b)]^2 }

= abs{sqrt(a) + sqrt(b)}

= sqrt(a) + sqrt(b).

Thus, the answer of this question is C

[1 / sqrt(5) ] * { [1 + sqrt(5)]/2 - [1 - sqrt(5)]/2 }

= [1 / sqrt(5)] * { [1 + sqrt(5) -1 + sqrt(5)] / 2}

= [1 / sqrt(5)] * { [2 * sqrt(5)] / 2}

= [1 / sqrt(5)] * sqrt(5)

= 1

Thus, answer is B

question 2:

Note that a>0 and b>0 , then sqrt(ab) = sqrt(a)*sqrt(b), sqrt(a)>0, sqrt(b)>0, and [sqrt(a) + sqrt(b)]>0.

We have:

sqrt{a + b + 2*sqrt(ab)}

= sqrt{ [sqrt(a)]^2 + 2*sqrt(a)*sqrt(b) + [sqrt(b)]^2 }

= sqrt{ [sqrt(a) + sqrt(b)]^2 }

= abs{sqrt(a) + sqrt(b)}

= sqrt(a) + sqrt(b).

Thus, the answer of this question is C

lmtuan

10-29-2004, 07:07 AM

Oh, I am sorry. I has a typo.

question 1

[1 / sqrt(5) ] * { [1 + sqrt(5)]/2 - [1 - sqrt(5)]/2 }

= [1 / sqrt(5)] * { [1 + sqrt(5) -1 + sqrt(5)] / 2}

= [1 / sqrt(5)] * { [2 * sqrt(5)] / 2}

= [1 / sqrt(5)] * sqrt(5)

= 1

Thus, answer is C

question 2:

Note that a>0 and b>0 , then sqrt(ab) = sqrt(a)*sqrt(b), sqrt(a)>0, sqrt(b)>0, and [sqrt(a) + sqrt(b)]>0.

We have:

sqrt{a + b + 2*sqrt(ab)}

= sqrt{ [sqrt(a)]^2 + 2*sqrt(a)*sqrt(b) + [sqrt(b)]^2 }

= sqrt{ [sqrt(a) + sqrt(b)]^2 }

= abs{sqrt(a) + sqrt(b)}

= sqrt(a) + sqrt(b).

Thus, the answer of this question is C

question 1

[1 / sqrt(5) ] * { [1 + sqrt(5)]/2 - [1 - sqrt(5)]/2 }

= [1 / sqrt(5)] * { [1 + sqrt(5) -1 + sqrt(5)] / 2}

= [1 / sqrt(5)] * { [2 * sqrt(5)] / 2}

= [1 / sqrt(5)] * sqrt(5)

= 1

Thus, answer is C

question 2:

Note that a>0 and b>0 , then sqrt(ab) = sqrt(a)*sqrt(b), sqrt(a)>0, sqrt(b)>0, and [sqrt(a) + sqrt(b)]>0.

We have:

sqrt{a + b + 2*sqrt(ab)}

= sqrt{ [sqrt(a)]^2 + 2*sqrt(a)*sqrt(b) + [sqrt(b)]^2 }

= sqrt{ [sqrt(a) + sqrt(b)]^2 }

= abs{sqrt(a) + sqrt(b)}

= sqrt(a) + sqrt(b).

Thus, the answer of this question is C

vn_snoopy

10-30-2004, 02:22 AM

Just apply (x+y)^2 = x^2 + y^2 + 2xy with x = sqrt(a) > 0 and y = sqrt(b) > 0. ;)

In this case, since x and y are both positive (hence the sum x + y), we can take the square root of (x + y)^2 as (x + y). And... so on ... :)

In this case, since x and y are both positive (hence the sum x + y), we can take the square root of (x + y)^2 as (x + y). And... so on ... :)

calm_J

11-06-2004, 11:04 AM

hi guys ,

the complete answer

the complete answer

sabna

11-08-2004, 11:53 AM

wow..nice

appreciable..!!!!!!

prayers

sabna

appreciable..!!!!!!

prayers

sabna

777

11-08-2004, 08:14 PM

1) When I did the subtraction, I got--

(2sqrt5/2)1/sqrt5 = 2/2 = 1

2) Square both sides and they're equal

(2sqrt5/2)1/sqrt5 = 2/2 = 1

2) Square both sides and they're equal