Important question plzzzzzzz help

**sscgeek** · June 13th 2013, 01:03 PM

If a = (3^1/2 + 2^1/2)^-3 and b = ((3^1/2 - 2^1/2)^-3 then the value of (a+1)^-1 + ( b+1)^-1 is

a)48 root 2

b)50 root 3

c) 1

d) 5

PLZZZZZZZZZZZZZZZZZZ EXPLAIN

**ebaines** · June 13th 2013, 02:31 PM

Not sure if there's an easier way to do this, but if $a = \frac 1 {(\sqrt 3 + \sqrt 2)^3}$ then multiplying it out yields

$a = \frac 1 {9 \sqrt 3 + 11 \sqrt 2}$

Similarly

$b = \frac 1 {9 \sqrt 3 - 11 \sqrt 2}$

For ease let's use notation $x = 9 \sqrt 3$ and $y = 11 \sqrt 3$ , so that:

$a = \frac 1 {x + y}; \ b = \frac 1 {x-y}$

Now:

$\frac 1 {a+ 1} + \frac 1 {b+1} = \frac 1 {\frac 1 {x+y} + 1} + \frac 1 {\frac 1 {x-y} + 1} = \frac {x+y}{1 + x + y} + \frac {x-y}{ 1 + x -y}$

This gets a little messy, but it boils down to:

$\frac {2x + 2x^2 -2y^2}{1 + 2x +x^2 - y^2}$

To finish you just need to sub in the values for x and y and see what you get.

**Soroban** · June 13th 2013, 05:09 PM

Hello, sscgeek!

$a \:=\: (\sqrt{3}+\sqrt{2})^{-3},\;\;b \:=\:(\sqrt{3}- \sqrt{2})^{-3}$

$\text{Find: }\:\frac{1}{a+1} + \frac{1}{b+1}$

. . . $(a) \;48\sqrt{2} \qquad (b)\;50\sqrt{3} \qquad (c)\;1 \qquad (d)\;5$

$\frac{1}{a+1} \;=\;\frac{1}{(\sqrt{3}+\sqrt{2})^{-3} + 1} \;=\;\frac{(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}+\sqrt{2})^3}$

$\frac{1}{b+1} \;=\;\frac{1}{(\sqrt{3}-\sqrt{2})^{-3} + 1} \;=\;\frac{(\sqrt{3}-\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3}$

$\frac{1}{a+1} + \frac{1}{b+1} \;=\; \frac{(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}+\sqrt{2})^3} + \frac{(\sqrt{3}-\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3}$

. . . . . . . . . . $=\;\frac{(\sqrt{3}+\sqrt{2})^3\left[1 + (\sqrt{3}-\sqrt{2})^3\right] + (\sqrt{3}-\sqrt{2})^3\left[1 (\sqrt{3}+\sqrt{2})^3\right]} {\left[1+(\sqrt{3}+\sqrt{2})^3\right]\left[1+(\sqrt{3}-\sqrt{2})^3\right]}$

. . . . . . . . . . $=\;\frac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3(\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3(\sqrt{3}-\sqrt{2})^3}$

. . . . . . . . . . $=\;\frac{(\sqrt{3}+\sqrt{2})^3 + (3-2)^3 + (\sqrt{3}-\sqrt{2})^3 + (3-2)^3}{1 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3 + (3-2)^3}$

. . . . . . . . . . $=\;\frac{(\sqrt{3}+\sqrt{2})^3 + 1 + (\sqrt{3}-\sqrt{2})^3 + 1}{1 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3 + 1}$

. . . . . . . . . . $=\;\frac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 + 2}{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 + 2}$

. . . . . . . . . . $=\; 1\;\text{ . . . answer (c)}$

**Idea** · June 13th 2013, 09:25 PM

(a+1)^-1 + ( b+1)^-1 = (a + b + 2) / (a + b + ab +1)
and ab=1

**ibdutt** · June 13th 2013, 10:11 PM

Originally Posted by ebaines

Not sure if there's an easier way to do this, but if $a = \frac 1 {(\sqrt 3 + \sqrt 2)^3}$ then multiplying it out yields

$a = \frac 1 {9 \sqrt 3 + 11 \sqrt 2}$

Similarly

$b = \frac 1 {9 \sqrt 3 - 11 \sqrt 2}$

For ease let's use notation $x = 9 \sqrt 3$ and $y = 11 \sqrt 3$ , so that:

$a = \frac 1 {x + y}; \ b = \frac 1 {x-y}$

Now:

$\frac 1 {a+ 1} + \frac 1 {b+1} = \frac 1 {\frac 1 {x+y} + 1} + \frac 1 {\frac 1 {x-y} + 1} = \frac {x+y}{1 + x + y} + \frac {x-y}{ 1 + x -y}$

This gets a little messy, but it boils down to:

$\frac {2x + 2x^2 -2y^2}{1 + 2x +x^2 - y^2}$

To finish you just need to sub in the values for x and y and see what you get.

Important question plzzzzzzz help-14-jun-13.png

**sscgeek** · June 13th 2013, 10:15 PM

HI BUDDY .CAN U PLZ EXPLAIN HW ab becomes 1

**Idea** · June 13th 2013, 10:23 PM

$\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1$

**sscgeek** · June 13th 2013, 11:54 PM

thanks buddy

**ibdutt** · June 14th 2013, 12:38 AM

Remember (a+b)(a-b)=a^2-b^2
Just apply this formula and you will get the result. for we have ab = (9sqrt3)^2-(11sqrt2)^2= 81*3 -121*2=243-242=1
Ok

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