Maths thread (1 Viewer)

leehuan · Jan 29, 2016

Lol you had a maths thread?

Maths questions you can just throw into the maths section of the forum without fear, since prelim maths gets reexamined in the HSC

drsabz101 · Jan 29, 2016

yeah It's okay

Green Yoda · Jan 29, 2016

I stumbled upon this question and have no idea how to do it.
Express in standard scientific notation form:
The number of cm^2 in 5 hectares, given that 1 hectare = 10^4m^2

Drsoccerball · Jan 29, 2016

Rathin said:
I stumbled upon this question and have no idea how to do it.
Express in standard scientific notation form:
The number of cm^2 in 5 hectares, given that 1 hectare = 10^4m^2

Just remember if going for m^2 to cm^2 you dont multiply by 100 but rather 100^2.

InteGrand · Jan 29, 2016

Rathin said:
I stumbled upon this question and have no idea how to do it.
Express in standard scientific notation form:
The number of cm^2 in 5 hectares, given that 1 hectare = 10^4m^2

$\noindent Note that $1 \text{ m} = 100\text{ cm}=10^2 \text{ cm}$. Hence (squaring both sides) $1 \text{ m}^2 = 10^4 \text{ cm}^2$ (as $1^2 = 1$ and $\left(10^2 \right)^2 = 10^4$). Now,$

$\begin{align*}5\text{ ha}&= 5 \times 10^4 \text{ m}^2 \quad (\text{as we are told that }1 \text{ ha} = 10^4 \text{ m}^2)\\ &= 5 \times 10^4 \times 10^4 \text{ cm}^2 \quad (\text{using the result we derived above, i.e. that }1 \text{ m}^2 = 10^4 \text{ cm}^2) \\ &= 5 \times 10^{8} \text{ cm}^2 \quad (\text{index laws}).\end{align*}$

$\noindent Hence the answer is $5\times 10^8$.$

$\noindent As you can see, the trick is basically to treat the units as though they are variables like $x$ and $y$, so we can square both sides of an equation involving them to obtain relationships between square units (or even cube both sides to get a relationship for cubic units for volume), or replace them with equalities we know for them in terms of other units etc. (E.g. squaring m makes it become $\text{m}^2$ (square-meters), just as though the m was just a pronumeral.)$

drsabz101 · Jan 29, 2016

OKAY so I have more questions :

How do I show an equation in this case x^2-4x+1 has 2 real and distinct solutions. Don't I just use the discriminant b^2-4ac>0?

InteGrand · Jan 29, 2016

drsabz101 said:
OKAY so I have more questions :

How do I show an equation in this case x^2-4x+1 has 2 real and distinct solutions. Don't I just use the discriminant b^2-4ac>0?

Yes.

drsabz101 · Jan 29, 2016

Ok cool

drsabz101 · Jan 29, 2016

How do I solve x - (4 square root x-1 ) +2=0.
do I sub a value for the square root of x-1.

DatAtarLyfe · Jan 29, 2016

I would let u=root of x-1 then go from there
Just remember to sub it back in at the end

InteGrand · Jan 29, 2016

drsabz101 said:
How do I solve x - (4 square root x-1 ) +2=0.
do I sub a value for the square root of x-1.

$\noindent We have $x-4\sqrt{x-1}+2=0$ (note that we must have $x\geq 1$ as the expression under the square root needs to be non-negative). Rearranging, we have $x+2 = 4\sqrt{x-1}$ (note that this implies that $x+2\geq 0$, i.e. $x\geq -2$. But we had $x\geq 1$ anyway, so this restriction of $x\geq -2$ was already implied). Squaring both sides, $x^2 + 4x + 4 = 16(x-1)\Longleftrightarrow x^2 + 4x + 4 = 16x - 16 \Longleftrightarrow x^2 - 12x + 20 = 0$. So $(x-2)(x-10) = 0$. So the solutions are $x=2,\,10$ (and both of these are permissible as they both satisfy the restriction $x\geq 1$).$

InteGrand · Jan 29, 2016

drsabz101 said:
see i did that and i got x minus the 4 square root u +2 =0 but what can i do next to this equation?

$\noindent If we take DatAtarLyfe's approach of substituting $u=x-1$, we need to replace all $x$'s in the original equations with $u$. Since $u=x-1$, we have $x=u+1$, so our equation becomes $(u+1)-4\sqrt{u}+2 = 0\Longleftrightarrow u - 4\sqrt{u} + 3 = 0$. We would then solve this for $u$ and then get the solutions for $x$ using $x=u+1$.$

leehuan · Jan 29, 2016

InteGrand said:
$\noindent If we take DatAtarLyfe's approach of substituting $u=x-1$, we need to replace all $x$'s in the original equations with $u$. Since $u=x-1$, we have $x=u+1$, so our equation becomes $(u+1)-4\sqrt{u}+2 = 0\Longleftrightarrow u - 4\sqrt{u} + 3 = 0$. We would then solve this for $u$ and then get the solutions for $x$ using $x=u+1$.$

He let u=sqrt(x+1) ---> u^2=x+1

Reduced the equation to something more obviously a quadratic.

drsabz101 · Jan 29, 2016

but then how would you solve for u from the equation ; u minus the square root of 4 +3=0

Drsoccerball · Jan 29, 2016

drsabz101 said:
but then how would you solve for u from the equation ; u minus the square root of 4 +3=0

$u- \sqrt{4} +3 = 0 ?$

$u= \sqrt{4} -3$

lol ?

drsabz101 · Jan 29, 2016

sorry haha wrote that wrong :/

it was u minus 4 square root u +3=0

Drsoccerball · Jan 29, 2016

drsabz101 said:
sorry haha wrote that wrong :/

it was u minus 4 square root u +3=0

pls learn Latex...

Here's one I made (No year 11 + help pls)
$Solve for x$

$4x^4+6x^3+3x^2+6x+4=0$

drsabz101 · Jan 29, 2016

Ignore the question I asked (solved it)

Latex seems quite complicated looking at that tutorial

Drsoccerball · Jan 29, 2016

drsabz101 said:
sorry haha wrote that wrong :/

it was u minus 4 square root u +3=0

But it's

$u -\sqrt{u}+3= 0$

$u-3 = \sqrt{u}$

$(u-3)^2=u$

$u^2-7u +9 =0$

leehuan · Jan 29, 2016

drsabz101 said:
Ignore the question I asked (solved it)

Latex seems quite complicated looking at that tutorial

Just use this. http://s1.daumcdn.net/editor/fp/service_nc/pencil/Pencil_chromestore.html

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