• Aug 2nd 2009, 01:19 PM
alexl13
I have no clue how to do these, and the book I have does not help at all.

$\displaystyle (\sqrt{3} + \sqrt{5})^2$
On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15}$

Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5})$

Solve.$\displaystyle \sqrt[3] {x-7}=2$

Solve$\displaystyle \sqrt{2x+1}=2x-5$
• Aug 2nd 2009, 01:25 PM
e^(i*pi)
Quote:

Originally Posted by alexl13
I have no clue how to do these, and the book I have does not help at all.

$\displaystyle (\sqrt{3} + \sqrt{5})^2$
On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15}$ e^(i*pi): you need only combine those radicals like you would with rational numbers and factorise

Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5})$

Solve.$\displaystyle \sqrt[3] {x-7}=2$

Solve$\displaystyle \sqrt{2x+1}=2x-5$

1. Use FOIL as normal:

$\displaystyle (\sqrt{3} + \sqrt{5})^2 = 3 + 2\sqrt{3}\sqrt{5} + 5 = 8 + 2\sqrt{15} = 2(4+\sqrt{15})$

2. It's the same as above but with different numbers

3. Cube to remove the radical:

$\displaystyle x-7 = 2^3 = 8$ - easy to see that x=15 (Wink)

4. Square to remove the radical, use FOIL on the RHS and solve the resulting quadratic
• Aug 2nd 2009, 01:25 PM
skeeter
Quote:

Originally Posted by alexl13
I have no clue how to do these, and the book I have does not help at all.

$\displaystyle (\sqrt{3} + \sqrt{5})^2$
On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15} \textcolor{red}{= 8 + 2\sqrt{15}}$ ... that's all.

Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5}) \textcolor{red}{= 19 + 2\sqrt{90} = 19 + 3\sqrt{10}}$

Solve.$\displaystyle \sqrt[3] {x-7}=2$

cube both sides ...

$\displaystyle \textcolor{red}{x-7 = 8}$

$\displaystyle \textcolor{red}{x = 15}$

Solve$\displaystyle \sqrt{2x+1}=2x-5$

square both sides, then solve the resulting quadratic ... don't forget to check your results in the original equation. go ahead, try it.

.
• Aug 2nd 2009, 01:26 PM
VonNemo19
Quote:

Originally Posted by alexl13
I have no clue how to do these, and the book I have does not help at all.

$\displaystyle (\sqrt{3} + \sqrt{5})^2$
On this one i got stuck after $\displaystyle 8+\sqrt{15}+\sqrt{15}$

Multiply.$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5})$

Solve.$\displaystyle \sqrt[3] {x-7}=2$

Solve$\displaystyle \sqrt{2x+1}=2x-5$

FOIL

$\displaystyle (\sqrt{3}+\sqrt{5})^2=(\sqrt{3})^2+\overbrace{\sqr t{3}\sqrt{5}+\sqrt{3}\sqrt{5}}^{\text{like terms}}+(\sqrt{5})^2$

$\displaystyle =3+2\sqrt{3}\sqrt{5}+5=3+2\sqrt{15}+5=8+2\sqrt{15}$
• Aug 2nd 2009, 01:30 PM
e^(i*pi)
Quote:

Originally Posted by skeeter
$\displaystyle (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5}) \textcolor{red}{= 19 + 2\sqrt{90} = 19 + 3\sqrt{10}}$

Where did you get get 90 from in Q2? I got $\displaystyle 19+\sqrt{70}$ with $\displaystyle \sqrt{70} = \sqrt{14}\sqrt{5}$
• Aug 2nd 2009, 01:48 PM
alexl13
Quote:

Originally Posted by e^(i*pi)

4. Square to remove the radical, use FOIL on the RHS and solve the resulting quadratic

I got $\displaystyle 2x+1=4x^2-20x=25$
but what confuses me is the exponent.
Would I make is a square root so that it would be 4x?
• Aug 2nd 2009, 01:53 PM
skeeter
Quote:

Originally Posted by e^(i*pi)
Where did you get get 90 from in Q2? I got $\displaystyle 19+\sqrt{70}$ with $\displaystyle \sqrt{70} = \sqrt{14}\sqrt{5}$

my mistake ... i was thinking 5 times 18 for some reason.
• Aug 2nd 2009, 01:56 PM
VonNemo19
Quote:

Originally Posted by alexl13
I got $\displaystyle 2x+1=4x^2-20x=25$
but what confuses me is the exponent.
Would I make is a square root so that it would be 4x?

$\displaystyle (\sqrt{2x+1})^2=(2x-5)^2$

$\displaystyle 2x+1=4x^2-10x+25$

$\displaystyle 4x^2-12x+24=0$

$\displaystyle x^2-3x+6=0$

$\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(6)}}{2(1)}$
• Aug 3rd 2009, 08:12 AM
e^(i*pi)
Quote:

Originally Posted by alexl13
I got $\displaystyle 2x+1=4x^2-20x=25$
but what confuses me is the exponent.
Would I make is a square root so that it would be 4x?

No you can solve the quadratic using factorising, completing the square or the quadratic equation

$\displaystyle (\sqrt{2x+1})^2=(2x-5)^2$

$\displaystyle 2x+1 = 4x^2 - 20x + 25$

$\displaystyle 4x^2 - 22x +25 = 0$

$\displaystyle x = \frac{22 \pm \sqrt{(-22)^2 - 4(4)(25)}}{2(4)}$

As $\displaystyle b^2-4ac > 0$ there will be two, distinct real solutions