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
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:
Looking at the LHS:
Move the a²b² over to the other side:
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
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)
Looking at we can see that 4q² is a factor so we can remove that to give
If we reintroduce a and b for p and q we get a solution of
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