How to solve a quadratic in a non quadratic form

**ric115** · May 4th 2009, 11:27 AM

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

**e^(i*pi)** · May 4th 2009, 11:31 AM

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

What is your actual equation? Remember there needs to be an equals sign in it

I would get everything on one side and then use the quadratic formula:

**ric115** · May 4th 2009, 11:44 AM

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

**e^(i*pi)** · May 4th 2009, 12:59 PM

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

**ric115** · May 4th 2009, 02:38 PM

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

**e^(i*pi)** · May 4th 2009, 03:50 PM

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)

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

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

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

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

**ric115** · May 5th 2009, 12:02 PM

Thank you very much that makes much more sense

Richard

**e^(i*pi)** · May 5th 2009, 12:07 PM

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

**ric115** · May 5th 2009, 12:21 PM

yeah got it

cheers again

**e^(i*pi)** · May 5th 2009, 12:25 PM

Originally Posted by ric115

yeah got it

cheers again

In that case it was a test to see if you spotted it

**ric115** · May 6th 2009, 09:52 AM

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)

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

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

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

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)

I hope you can follow what i ve written lol

Cheers again

Richard

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