1. ## How to solve a quadratic in a non quadratic form

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

2. 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:

3. 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

4. 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:

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

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

Move the a²b² over to the other side:

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

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

5. 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

6. 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)

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

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

$\displaystyle s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}$$\displaystyle = \frac{2q(-2 {\color{red}p}\pm \sqrt{p^2-1})}{2} = q(-2{\color{red}p}\pm \sqrt{p^2-1}) If we reintroduce a and b for p and q we get a solution of \displaystyle s = b(-2{\color{red}a}\pm \sqrt{a^2-1}) From there you put in your values of a and b 7. Thank you very much that makes much more sense Richard 8. Originally Posted by ric115 Thank you very much that makes much more sense Richard Just to let you know I made a mistake in my previous work but is now corrected in red 9. yeah got it cheers again 10. Originally Posted by ric115 yeah got it cheers again In that case it was a test to see if you spotted it 11. Originally Posted by e^(i*pi) You can use it because b^2 is a constant in terms of s (ie: there is no s term there) Let a =p and b=q s² + 2abs + b² = s² + 2pqs + q^2 = 0 (this will make the next step easier to understand), In this equation we know that a=1, b = 2pq and c=q^2 (the coefficients of s^2, s and constant respectively) \displaystyle s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm \sqrt{4p^2q^2 - 4q^2}}{2} Looking at \displaystyle \sqrt{4p^2q^2 - 4q^2} we can see that 4q² is a factor so we can remove that to give \displaystyle \sqrt{4q^2} \sqrt{p^2-1} = 2q \sqrt{p^2-1} \displaystyle s = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-2pq \pm 2q\sqrt{p^2-1}}{2}$$\displaystyle = \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

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

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