when applying a formula should i open brackets?

Question

I need to solve the following equation but am not sure how to, i would out that i would need to use the quadratic formula. In the question i had to rearrange so the subject is 'b' but before i apply formula i need to put it in the order 'ax^2+bx+c=0'or some people write it as 'ax*2+bx+c=0'. In order for me to even do that i need to know if i would have to open the brackets in this equation or not?

And can you please tell me what the substitution in the formula is going to be? (formula : ax^2+bx+c=0)

Equation : a = b(b+(r/nx)^2)
a equals b multiplied by b plus r divided by nx (r/nx = one part) whole squared

Answer 1

Since you are solving for b, you will get after expanding and transposing:



(nx)^2 b^2 + (r^2) b - a (nx)^2 = 0



So, in the quadratic formula for AX^2 + BX + C = 0, you will have



A = (nx)^2, B = r^2, and C = -a (nx)^2.

Answer 2

r/nx as shown AS PER PEDMAS is rx/n ...as part of (rx/n)^2 [you want to say [r/(nx)]^2 when writing on one line WHICH IS WHAT A COMPUTER READS as input {and your English description is also wrong.} To wit:



a = b*b + (r/n)*x ... I cannot translate into the formula 'whole squared' {even though I know what you mean



use the terms QUANTITY, and END QUANTITY to note delimiters {emphasis mine.}



I can get you to -(b^2) = b * {[r/(nx)]^2} - a ... but that does not appear to be the correct path

try 0 = b^2 + {ditto} -a



beyond that, my math skills are not sharp this morning

Answer 3

I think in the middle you drifed off because when you expand the brackets you you have to multiply b^2 by b which makes it b^3 and imposible to solve without the quadratic formula

Answer 4

You could try substituting "z" for (r/nx).

Then it becomes a = b ( b + ( z^2))

... which is a = b^2 + bz^2

... which is 0 = bz^2 + b^2 - a

... that's the same format as Ax^2 + Bx + C = 0, except with different values for a, b, and c.



... so to make "b" the subject, we have

0 = b ( z^2 + 1) - a

... which is the same as

b ( z^2 + 1) = a

... so we get

b = a / (z^2 + 1)

... then substitute for z

b = a / ((r/nx)^2 + 1)