# Thread:

1. ## Radicals

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

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

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

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

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

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

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

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

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

Solve $\sqrt{2x+1}=2x-5$
1. Use FOIL as normal:

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

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

3. Cube to remove the radical:

$x-7 = 2^3 = 8$ - easy to see that x=15

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

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

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

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

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

cube both sides ...

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

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

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

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

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

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

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

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

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

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

$=3+2\sqrt{3}\sqrt{5}+5=3+2\sqrt{15}+5=8+2\sqrt{15}$

5. Originally Posted by skeeter
$<br /> (\sqrt{14}+\sqrt{5})(\sqrt{14}+\sqrt{5}) \textcolor{red}{= 19 + 2\sqrt{90} = 19 + 3\sqrt{10}}<br />$
Where did you get get 90 from in Q2? I got $19+\sqrt{70}$ with $\sqrt{70} = \sqrt{14}\sqrt{5}$

6. Originally Posted by e^(i*pi)

4. Square to remove the radical, use FOIL on the RHS and solve the resulting quadratic
I got $2x+1=4x^2-20x=25$
but what confuses me is the exponent.
Would I make is a square root so that it would be 4x?

7. Originally Posted by e^(i*pi)
Where did you get get 90 from in Q2? I got $19+\sqrt{70}$ with $\sqrt{70} = \sqrt{14}\sqrt{5}$
my mistake ... i was thinking 5 times 18 for some reason.

8. Originally Posted by alexl13
I got $2x+1=4x^2-20x=25$
but what confuses me is the exponent.
Would I make is a square root so that it would be 4x?
$(\sqrt{2x+1})^2=(2x-5)^2$

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

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

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

$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(6)}}{2(1)}$

9. Originally Posted by alexl13
I got $2x+1=4x^2-20x=25$
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

$<br /> (\sqrt{2x+1})^2=(2x-5)^2<br />$

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

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

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

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