# Thread: 3 hard algebra problems!

1. ## 3 hard algebra problems!

Just can't figure out these 3 problems:

Thanks.

2. Originally Posted by xxravenxx
Just can't figure out these 3 problems:

Thanks.
HI

For the first problem ,

$\displaystyle x^2+ax-b=0$ its roots are a and c

$\displaystyle a+c=-a$ ---1

$\displaystyle ac=-b$ ----2

$\displaystyle x^2+cx+d=0$ its roots are b and d ,

$\displaystyle b+d=-c$ --- 3

$\displaystyle bd=d$ ---4

From 4 , $\displaystyle bd=d\Rightarrow b=1$

From 1 , $\displaystyle 2a=-c$

At 2 , $\displaystyle a(-2a)=-b$

$\displaystyle 2a^2=b$

$\displaystyle 2a^2=1$

$\displaystyle a=\pm\frac{1}{\sqrt{2}}$

At 1 , $\displaystyle 2(\pm\frac{1}{\sqrt{2}})=-c$

$\displaystyle c=2(\mp\frac{1}{\sqrt{2}})$

$\displaystyle c=\mp\sqrt{2}$

Then , from 2 ,

$\displaystyle 1+d=-c$

$\displaystyle 1+d=\pm\sqrt{2}$

$\displaystyle d=-1\pm\sqrt{2}$

HI

For the first problem ,

$\displaystyle x^2+ax-b=0$ its roots are a and c

$\displaystyle a+c=-a$ ---1

$\displaystyle ac=-b$ ----2

$\displaystyle x^2+cx+d=0$ its roots are b and d ,

$\displaystyle b+d=-c$ --- 3

$\displaystyle bd=d$ ---4

From 4 , $\displaystyle bd=d\Rightarrow b=1$

From 1 , $\displaystyle 2a=-c$

At 2 , $\displaystyle a(-2a)=-b$

$\displaystyle 2a^2=b$

$\displaystyle 2a^2=1$

$\displaystyle a=\pm\frac{1}{\sqrt{2}}$

At 1 , $\displaystyle 2(\pm\frac{1}{\sqrt{2}})=-c$

$\displaystyle c=2(\mp\frac{1}{\sqrt{2}})$

$\displaystyle c=\mp\sqrt{2}$

Then , from 2 ,

$\displaystyle 1+d=-c$

$\displaystyle 1+d=\pm\sqrt{2}$

$\displaystyle d=-1\pm\sqrt{2}$

Thank you for giving the solution to Problem 4, however, could you please explain how you got that sign? I don't understand how you got from this:

At 1 , $\displaystyle 2(\pm\frac{1}{\sqrt{2}})=-c$

to this:

$\displaystyle c=2(\mp\frac{1}{\sqrt{2}})$

I have never come across that symbol before. Care to explain what it means please?

And any one have solutions for the other two questions?

4. Originally Posted by xxravenxx
Thank you for giving the solution to Problem 4, however, could you please explain how you got that sign? I don't understand how you got from this:

At 1 , $\displaystyle 2(\pm\frac{1}{\sqrt{2}})=-c$

to this:

$\displaystyle c=2(\mp\frac{1}{\sqrt{2}})$

I have never come across that symbol before. Care to explain what it means please?

And any one have solutions for the other two questions?
$\displaystyle \pm\frac{1}{sqrt{2}}$ means $\displaystyle \frac{1}{sqrt{2}}$ or $\displaystyle -\frac{1}{\sqrt{2}}$

Take for example $\displaystyle x^2=4$

$\displaystyle x=\pm 2$

That means x can be 2 or -2 as well .

Does that clear things up for you ?

$\displaystyle \pm\frac{1}{sqrt{2}}$ means $\displaystyle \frac{1}{sqrt{2}}$ or $\displaystyle -\frac{1}{\sqrt{2}}$

Take for example $\displaystyle x^2=4$

$\displaystyle x=\pm 2$

That means x can be 2 or -2 as well .

Does that clear things up for you ?
Yea I know that symbol its just that upside down one that baffles me?

6. Originally Posted by xxravenxx
Yea I know that symbol its just that upside down one that baffles me?
When you have , $\displaystyle \pm 2$

$\displaystyle -(\pm 2)=\mp 2$

$\displaystyle 2 \pm 3 = 5 or -1$

$\displaystyle 2\mp 3 = -1 or 5$

When you have , $\displaystyle \pm 2$

$\displaystyle -(\pm 2)=\mp 2$

$\displaystyle 2 \pm 3 = 5 or -1$

$\displaystyle 2\mp 3 = -1 or 5$

Alright. Thanks again.

Any clue on the others? :S

,

,

,

,

### hardest algebra problem

Click on a term to search for related topics.