• May 4th 2009, 11:27 AM
ric115
How to Solve a quadratic equation in a non quadratic form such as :

s² + 2abs + b² s = 4
a = 5
b = 2

Do I solve the solution by completing the square?

My tutor has advised me to substitute some values into the equation which I have done shown above, with these values it = 100

He said also to re arrange and maybe move the b² to the right hand side s² + 2abs = -b²

Do you know why he has advised me to do this?

As you can see im kinda hitting a brick wall with this quadratic so any help would be much appreciated :O)

Richard
• May 4th 2009, 11:31 AM
e^(i*pi)
Quote:

Originally Posted by ric115
How to Solve a quadratic equation in a non quadratic form such as :

s² + 2abs + b² s = 4
a = 5
b = 2

Do I solve the solution by completing the square?

My tutor has advised me to substitute some values into the equation which I have done shown above, with these values it = 100

He said also to re arrange and maybe move the b² to the right hand side s² + 2abs = -b²

Do you know why he has advised me to do this?

As you can see im kinda hitting a brick wall with this quadratic so any help would be much appreciated :O)

Richard

I would get everything on one side and then use the quadratic formula:
• May 4th 2009, 11:44 AM
ric115
This is the original equation sorry my message wasnt very clear.

s² + 2abs + b²

s = 4
a = 5
b = 2

My tutor has told me to move the b² to the right hand side

s² + 2abs = -b²

Could i substitue these values into the equation and have

16 + 80 + 4 = 100

Thanks

Richard
• May 4th 2009, 12:59 PM
e^(i*pi)
Quote:

Originally Posted by ric115
This is the original equation sorry my message wasnt very clear.

s² + 2abs + b²

s = 4
a = 5
b = 2

My tutor has told me to move the b² to the right hand side

s² + 2abs = -(b²)

Could i substitue these values into the equation and have

16 + 80 + 4 = 100

Thanks

Richard

Given what you said I'll assume it is equal to 0 then. If you move b² over to the right hand side as said you can complete the square:

$s^2 + 2abs = -b^2$

Looking at the LHS: $s^2+2abs = (s+ab)^2 - {\color{blue}a^2b^2}$

Move the a²b² over to the other side:

$(s+ab)^2 = a^2b^2-b^2 = b^2(a^2-1)$

Can you go on from there?

Personally I still think using the quadratic equation would be easier with a=1, b= 2ab, c=b^2 :p
• May 4th 2009, 02:38 PM
ric115
This may sound silly but if i was to use the quadratic equation what would s = ?
I didnt realise i could use the quadratic formula as i the equation had two square values

Cheers
Richard
• May 4th 2009, 03:50 PM
e^(i*pi)
Quote:

Originally Posted by ric115
This may sound silly but if i was to use the quadratic equation what would s = ?
I didnt realise i could use the quadratic formula as i the equation had two square values

Cheers
Richard

You can use it because b^2 is a constant in terms of s (ie: there is no s term there)

Let a =p and b=q

s² + 2abs + b² = s² + 2pqs + q^2 = 0 (this will make the next step easier to understand),

In this equation we know that a=1, b = 2pq and c=q^2 (the coefficients of s^2, s and constant respectively)

$
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm \sqrt{4p^2q^2 - 4q^2}}{2}$

Looking at $\sqrt{4p^2q^2 - 4q^2}$ we can see that 4q² is a factor so we can remove that to give $\sqrt{4q^2} \sqrt{p^2-1} = 2q \sqrt{p^2-1}$

$
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}$
$= \frac{2q(-2 {\color{red}p}\pm \sqrt{p^2-1})}{2} = q(-2{\color{red}p}\pm \sqrt{p^2-1})$

If we reintroduce a and b for p and q we get a solution of

$s = b(-2{\color{red}a}\pm \sqrt{a^2-1})$

From there you put in your values of a and b
• May 5th 2009, 12:02 PM
ric115
Thank you very much that makes much more sense

Richard
• May 5th 2009, 12:07 PM
e^(i*pi)
Quote:

Originally Posted by ric115
Thank you very much that makes much more sense

Richard

Just to let you know I made a mistake in my previous work but is now corrected in red
• May 5th 2009, 12:21 PM
ric115
yeah got it

cheers again
• May 5th 2009, 12:25 PM
e^(i*pi)
Quote:

Originally Posted by ric115
yeah got it

cheers again

In that case it was a test to see if you spotted it (Wink)
• May 6th 2009, 09:52 AM
ric115
Quote:

Originally Posted by e^(i*pi)
You can use it because b^2 is a constant in terms of s (ie: there is no s term there)

Let a =p and b=q

s² + 2abs + b² = s² + 2pqs + q^2 = 0 (this will make the next step easier to understand),

In this equation we know that a=1, b = 2pq and c=q^2 (the coefficients of s^2, s and constant respectively)

$
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm \sqrt{4p^2q^2 - 4q^2}}{2}$

Looking at $\sqrt{4p^2q^2 - 4q^2}$ we can see that 4q² is a factor so we can remove that to give $\sqrt{4q^2} \sqrt{p^2-1} = 2q \sqrt{p^2-1}$

$
s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}$
$= \frac{2q(-2 {\color{red}p}\pm \sqrt{p^2-1})}{2} = q(-2{\color{red}p}\pm \sqrt{p^2-1})$

If we reintroduce a and b for p and q we get a solution of

$s = b(-2{\color{red}a}\pm \sqrt{a^2-1})$

From there you put in your values of a and b

hi

Regarding the formula shown below i have gone through this but theres only one part i dont understand. the last bit of the equation. why do you put brackets around s = 2q(-2p ± √b² - 1) or should the formula be
s = 2q(-2pq ± √b² - 1)

http://www.mathhelpforum.com/math-he...00fcb925-1.gifhttp://www.mathhelpforum.com/math-he...fbf3afbe-1.gif

I hope you can follow what i ve written lol

Cheers again

Richard