prove !!!

• Jul 29th 2008, 07:27 PM
qweiop90
prove !!!
3<(13)^1/2< 4, deduce that 0<(13)^1/2 -3<1

b) [(13^1/2)+3]^4 + [(13^1/2)-3)]^4 =1904

c)1903<[(13^1/2)+3]^4 <1904

here is another question

expresss [(p^1/2 +q(r^1/2)]^2 in the form =a+b(c^1/2). without evaluating the square root or using table or calculator,show that 10^1/2 +2(2^1/2) <6
• Jul 29th 2008, 07:31 PM
Mathstud28
Quote:

Originally Posted by qweiop90
3<(13)^1/2< 4, deduce that 0<(13)^1/2 -3<1

b) [(13^1/2)+3]^4 + [(13^1/2)-3)]^4 =1904

c)1903<[(13^1/2)+3]^4 <1904

here is another question

expresss [(p^1/2 +q(r^1/2)]^2 in the form =a+b(c^1/2). without evaluating the square root or using table or calculator,show that 10^1/2 +2(2^1/2) <6

Square each term of the first problem

$\displaystyle 3\leq\sqrt{13}\leq{4}$

$\displaystyle \Rightarrow{9\leq{13}\leq{16}}$
• Jul 29th 2008, 07:43 PM
qweiop90
stil couldnt get

9<13<16

frm the question : (13^1/2+3)^4+(13^1/2-3)^4=1904

how do you get? i had squared, but i stil couldnt get
• Jul 29th 2008, 07:46 PM
Mathstud28
Quote:

Originally Posted by qweiop90
stil couldnt get

9<13<16

frm the question : (13^1/2+3)^4+(13^1/2-3)^4=1904

how do you get? i had squared, but i stil couldnt get

$\displaystyle 3\leq\sqrt{13}\leq{4}$

$\displaystyle \Rightarrow{3-3\leq\sqrt{13}-3\leq{1}}$

$\displaystyle \Rightarrow{0\leq\sqrt{13}-3\leq{1}}$

That is what you wanted right?
• Jul 29th 2008, 07:49 PM
qweiop90
the first question i already solved. yet, by using the same method, i stil couldnt prove the 2nd and 3rd question