Two surds problems - one involving simultaneous equations..

**struck** · August 7th 2008, 08:36 AM

1. Two sides of a triangle are 5sqrt(3) and 10sqrt(2) respectively, while the hypotenuse is unknown. Here I tried finding the hypotenuse using pythagoras theorem.

x ^ 2 = (5sqrt(3))^2 + (10sqrt(2)) ^ 2

The answer on should be x = 5sqrt(11) but I get 5sqrt(3) + 10

2. Two simultaneous equations are as follows:

Code:

7x - (3sqrt(5))y = 9sqrt(5)           (1)
(2sqrt(5)x + y = 34                     (2)

I tried rearranging (1) to y = (9sqrt(5) - 7x) / 3sqrt(5) but I cannot reach an answer at all

..

Please help..

P.S. Is there any tutorial on how to generate real maths symbols on this forum?

**Soroban** · August 7th 2008, 08:58 AM

Hello, struck!

1. Two sides of a right triangle are $5\sqrt{3}$ and $10\sqrt{2}$
Find the hypotenuse.

$x^2 \:= \<img src=$ 5\sqrt{3})^2 + (10\sqrt{2})^2 " alt="x^2 \:= \

5\sqrt{3})^2 + (10\sqrt{2})^2 " />

$x^2 \;=\;5^2(\sqrt{3})^2 + 10^2(\sqrt{2})^2$

$x^2 \;=\;25(3) + 100(2) \;=\;75 + 200 \;=\;275$

$x \;=\;\sqrt{275} \;=\;\sqrt{25\!\cdot\!11} \;=\;\boxed{5\sqrt{11}}$

$2)\;\;\begin{array}{cccc}7x - (3\sqrt{5})y &= &9\sqrt{5} & {\color{blue}[1]} \\<br /> (2\sqrt{5})x + y &=& 34 & {\color{blue}[2]} \end{array}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[2]} by }3\sqrt{5}\!: & 30x + 3\sqrt{5}\,y & = & 102\sqrt{5} \\<br /> \text{Add {\color{blue}[1]}:} & 7x - 3\sqrt{5}\,y &=& 9\sqrt{5} \end{array}$

And we have: . $37x \:=\:111\sqrt{5} \quad\Rightarrow\quad \boxed{x \:=\:3\sqrt{5}}$

Substitute into [2]: . $(2\sqrt{5})(3\sqrt{5}) + y \:=\:34 \quad\Rightarrow\quad 30 + y \:=\:34 \quad\Rightarrow\quad \boxed{y \:=\:4}$

**badgerigar** · August 7th 2008, 11:38 PM

P.S. Is there any tutorial on how to generate real maths symbols on this forum?

you can click on anyone's math symbols to see what they used to write something. Also, there is a branch of the forum where you can ask if you are having difficulty. This page might also come in handy

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