# Math Help - I need help with quadratics and indicies?

1. ## I need help with quadratics and indicies?

Factorise: x2+3xy+2y2

at the end instead of a number so I ended up with this:

= (x + 2y2)2 - (xy - 2y2)

Because

x2+ 4xy + 4y2 - xy - 2y2

= x2 + 3xy + 2y2

But it's not properly factoried so I'm unsure about it?

The next one was factorise: 4x2+ 2x + 1

I'm confused because
1 and 1 = 2 and 2 (*2) = 4 or 0 - not 2
4 and 1 (*4 and *1) = 5 or 3 - also not 2
and neither of these combinations work so I don't know what else to do?

And the last one is solve: (1/49)-1/2

I know that it means square root of 1/1/7 because the power1/2 = square root and the - means 1 over but 1/1/7 isn't a rational number and when I try turning it into a decemal it goes on forever and because it would be 1/0.143 (rounded) its still not a rational number so I am incredibly confused and would really apreciate help..? By the way I'm not aloud to use a calculator so could you please explain without one?

2. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
Factorise: x2+3xy+2y2

at the end instead of a number so I ended up with this:

= (x + 2y2)2 - (xy - 2y2)

Because

x2+ 4xy + 4y2 - xy - 2y2

= x2 + 3xy + 2y2

But it's not properly factoried so I'm unsure about it?

The next one was factorise: 4x2+ 2x + 1

I'm confused because
1 and 1 = 2 and 2 (*2) = 4 or 0 - not 2
4 and 1 (*4 and *1) = 5 or 3 - also not 2
and neither of these combinations work so I don't know what else to do?

And the last one is solve: (1/49)-1/2

I know that it means square root of 1/1/7 because the power1/2 = square root and the - means 1 over but 1/1/7 isn't a rational number and when I try turning it into a decemal it goes on forever and because it would be 1/0.143 (rounded) its still not a rational number so I am incredibly confused and would really apreciate help..? By the way I'm not aloud to use a calculator so could you please explain without one?
If in doubt, you can always complete the square on one of the variables...

\displaystyle \begin{align*} x^2 + 3\,x\,y + 2y^2 &= x^2 + 3\,x\,y + \left(\frac{3}{2}y\right)^2 - \left(\frac{3}{2}y\right)^2 + 2y^2 \\ &= \left(x + \frac{3}{2}y\right)^2 - \frac{9}{4}y^2 + \frac{8}{4}y^2 \\ &= \left(x + \frac{3}{2}y\right)^2 - \frac{1}{4}y^2 \\ &= \left(x + \frac{3}{2}y\right)^2 - \left(\frac{1}{2}y\right)^2 \\ &= \left(x + \frac{3}{2}y - \frac{1}{2}y\right)\left(x + \frac{3}{2}y + \frac{1}{2}y\right) \\ &= \left(x + y\right)\left(x + 2y\right) \end{align*}

3. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
Factorise: x2+3xy+2y2

at the end instead of a number so I ended up with this:

= (x + 2y2)2 - (xy - 2y2)
You should have (x+ 2y)2 as he first term. Squaring y2 will give you y4 that you don't want.

Because

x2+ 4xy + 4y2 - xy - 2y2

= x2 + 3xy + 2y2

But it's not properly factoried so I'm unsure about it?

The next one was factorise: 4x2+ 2x + 1

I'm confused because
1 and 1 = 2 and 2 (*2) = 4 or 0 - not 2
4 and 1 (*4 and *1) = 5 or 3 - also not 2
and neither of these combinations work so I don't know what else to do?

And the last one is solve: (1/49)-1/2

I know that it means square root of 1/1/7 because the power1/2 = square root and the - means 1 over but 1/1/7 isn't a rational number and when I try turning it into a decemal it goes on forever and because it would be 1/0.143 (rounded) its still not a rational number so I am incredibly confused and would really apreciate help..? By the way I'm not aloud to use a calculator so could you please explain without one?

4. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
Factorise: x2+3xy+2y2

at the end instead of a number so I ended up with this:

= (x + 2y2)2 - (xy - 2y2)

Because

x2+ 4xy + 4y2 - xy - 2y2

= x2 + 3xy + 2y2

But it's not properly factoried so I'm unsure about it?
First, that was a great way to decompose the quadratic. Kudos!

There is a way to do this that will work in most cases, but this one is easier to use the "guess method."

The coefficient on the $x^2$ is 1 and the coefficient of the $y^2$ is 2. So the possible factors of your expression must be of the form $(x \pm y)(x \pm 2y)$ where the signs on the y terms are the same.

Now just play around with $\pm$ until you get the correct answer.

If that's too hard to see then I can give you the longer (but more mechanical) way to do it. It's called the "ac" method.

-Dan

5. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
Factorise: x2+3xy+2y2

at the end instead of a number so I ended up with this:

= (x + 2y2)2 - (xy - 2y2)

Because

x2+ 4xy + 4y2 - xy - 2y2

= x2 + 3xy + 2y2

But it's not properly factoried so I'm unsure about it?

The next one was factorise: 4x2+ 2x + 1

I'm confused because
1 and 1 = 2 and 2 (*2) = 4 or 0 - not 2
4 and 1 (*4 and *1) = 5 or 3 - also not 2
and neither of these combinations work so I don't know what else to do?

And the last one is solve: (1/49)-1/2

I know that it means square root of 1/1/7 because the power1/2 = square root and the - means 1 over but 1/1/7 isn't a rational number and when I try turning it into a decemal it goes on forever and because it would be 1/0.143 (rounded) its still not a rational number so I am incredibly confused and would really apreciate help..? By the way I'm not aloud to use a calculator so could you please explain without one?
\displaystyle \begin{align*} 4x^2 + 2x + 1 \end{align*} does not factorise over the reals (though it does over the complex numbers). Checking the discriminant will prove this.

6. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
Factorise: x2+3xy+2y2
(following is "I did it my way"!):
Remove the y's:
x^2 + 3x + 2
Factor:
(x + 2)(x + 1)
Put the y's back in:
(x + 2y)(x + y)

7. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
And the last one is solve: (1/49)-1/2

I know that it means square root of 1/1/7 because the power1/2 = square root and the - means 1 over but 1/1/7 isn't a rational number and when I try turning it into a decemal it goes on forever and because it would be 1/0.143 (rounded) its still not a rational number so I am incredibly confused and would really apreciate help..? By the way I'm not aloud to use a calculator so could you please explain without one?
$49 = 7^2$

and
$x^{-y} = \left ( \frac{1}{x} \right )$

To get you started...
$\left ( \frac{1}{49} \right ) ^{-1/2} = \left ( \frac{1}{7^2} \right ) ^{-1/2}$

Can you finish from here? (By the way, you did get the answer: $\frac{1}{ \frac{1}{7} } = 7$ which is the correct answer.

-Dan

8. ## Re: I need help with quadratics and indicies?

Thank you so much!

9. ## Re: I need help with quadratics and indicies?

I'm still confused on the middle one, does it just not factorise?

10. ## Re: I need help with quadratics and indicies?

I'm sorry I've just clicked to that, so saying 1/1/7 would be the same as just saying 1 * 7, just like saying 1*1/7 is the same as saying 1/7.. That makes so much sence I just didn't click lol, thank you so much

11. ## Re: I need help with quadratics and indicies?

I've been trying to factorise it and the closest thing I got was this:

4x2+2x+1= (2x+√2x+1)2

I am really confuesed because it doesn't seem to factorise but that's what I'm being asked to do?

12. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
I've been trying to factorise it and the closest thing I got was this:
4x2+2x+1= (2x+√2x+1)2
I am really confuesed because it doesn't seem to factorise but that's what I'm being asked to do?
You were already told it cannot be factorized.
Where did you get that expression? Your math teacher?
Are you sure it's correct?
Would you be able to factorize it IF it was 4x^2 + 2x - 3 ?

13. ## Re: I need help with quadratics and indicies?

Originally Posted by Tabitha
I'm sorry I've just clicked to that, so saying 1/1/7 would be the same as just saying 1 * 7, just like saying 1*1/7 is the same as saying 1/7.. That makes so much sence I just didn't click lol, thank you so much
1/1/7 = .142857....
1/(1/7) = 7
Make sure you understand that...