Quadratic equation help (again!) please!!?

Question

good evening. PLease could i have some assistance in the workings of a quadratic equation. THanks

I have the equation:



((square root ((5.37 x 10^-10) x (0.0000135) + [Z])) x [Z]) = 1 x 10^-14

I would like to find the quadratic equation in [Z] ( which represents the concentration of protons in my experiment)

I have tried for an hour and cant do it, so any help greately appreciated. Thanks

Answer 1

It's NOT quadratic!

Let's cut through all the numerical details and get rid of the (far too many!) confusing parentheses and square brackets by writing your equation as:

z sqrt(a b + z) = c ..........(1)

This is much easier to handle --- the actual numerics can be entered later, if desired.

Squaring equation (1), we have:

z^2 (a b + z) = c^2, ..........(2)

or: z^3 + a b z^2 - c^2 = 0. ..........(3)

There's NO getting around it --- this is a CUBIC equation! Your search for a quadratic equation will prove in vain.

Live long and prosper.

P.S. Wal C (below) is generally very good on math questions. But even he was confused by your plethora of parentheses. (He only ended up with a quadratic equation by taking the surd sign completely outside all the remaining brackets when that can't in fact be done.)

Answer 2

To me this looks like a cubic equation.

Here's why. I simplified your equation by using e notation and by getting rid of parentheses when I can. I also did your multiplication of 5.37e-10 by 0.0000135.

square root (7.2495e-15 + Z)*Z = 1e-14

To get rid of the radical, square both sides.

(7.2495e-15 + Z)*Z^2 = 1e-28

Multiplying out,

7.2495e-15*Z^2 + Z^3 = 1e-28
Z^3+7.2495e-15*Z^2-1e-28 = 0

This equation could be really terrible to solve, because the extreme range of the quantities involved, and because it is a cubic equation. Some things that might help:

Substitute Z = 1/y. That puts it in the form y^3+py+q = 0, which is easier to solve.

-1e-28*y^3+7.2495e-15*y +1= 0 or
1e-28*y^3 - 7.2495e-15*y - 1 = 0

To deal with the ill-conditioning, try substituting y = x*10e9. (Since you are talking about protons (H+ ions) and chemistry, the substitution y = x*10e7 sticks out as a possibility here, since a concentration of 1e-7 H+ ions represents a pH of 7, but 1e9 is suggested by the e-28 that appears in the equation.)

0.1x^3 - 7.2495e-6*x - 1 = 0

or

x^3 - 7.2495e-10x - 10 = 0

The best thing to do now is to use Newton's Method. Since the middle term is so small, drop it. The resulting approximation is

x^3 = 10
or x = cbrt(10) ~ 2.1544

This should come close to your answer. Use this as your initial approximation and apply Newton's method and you will have your answer.

Also, your problem was confusing with its multitude of parentheses. Maybe you meant this equation instead:

square root ( (7.2495e-15 + Z)*Z) = 1e-14

with the Z inside the square root. Now squaring will result in

(7.2495e-15+Z)*Z = 1e-28

or

Z^2+7.2495e-15Z - 1e-28 = 0

Substitute Z = 1e14x. Note the 14 - the range of pH values. To me that makes this interpretation of your equation more likely.

x^2 + 0.72495*x - 1e-28 = 0

Solve this by the quadratic equation, but remember that doing this will lose significant digits - maybe this is where you had problems.

x = (-0.72495 + sqrt(0.72495^2 + 4e-28)) / 2

You will need a super-precision calculator for this because this includes the sum of two nearly equal quantities. Another way is to use the approximation:

sqrt(x^2+a) = (x^2+a)^(1/2) = x + a/2x + O(a^2)

which can be derived from the binomial formula.

This gives

(0.72495 + (1/2)*(-4e-28)/0.72495-0.72495)/2
= 2.7588*e-28.

This is a very small quantity, but if you drop 1e-14 from your equation, the resulting equation has 0 for a root, so this confirms that this root will be tiny.

Hope all this helps with what I regard as a difficult problem to express and to solve.

Answer 3

((√ ((5.37 x 10^-10) x (0.0000135) + [Z])) x [Z]) = 1 x 10^-14

ie√ ((5.37 x 10^-10) x (0.0000135)x [Z] + [Z]²) = 1 x 10^-14

Square both sides

So [Z]² + 7.2495 x 10^-15 x [Z] - 1 x 10^-28 = 0

So by quadratic formula:

[Z] =
½(-7.2495 x 10^-15 ± √[(7.2495 x 10^-15)² - 4 (1) (- 1 x 10^-28)

= ½(-7.2495 x 10^-15 ± √[4.5255525025 x 10^-28])

≈ ½(-7.2495 x 10^-15 ± 2.1273346 x 10^-14)

Given that [Z] should be a positive quantity the negative root is ignored

So [Z] ≈ ½(-7.2495 x 10^-15 + 2.1273346 x 10^-14)
≈ 7.012 x 10^(-15)

Answer 4

It's a cubic equation, not a quadratic:

[Z]√(5.37E-10*.0000135+[Z])
=1E-14

(I think you can guess what E means.)

Squaring both sides and multiplying two numbers together (surely you could have done that!):
[Z]²(7.2495E-15 +[Z])=1E-28

Multiplying out brackets and rearranging:
[Z]³ + 7.2495E-15[Z]²-1E-28=0

We now go to the cubic equation calculator (see reference below but many calculators have this function) and enter:

A=1; B=7.2495E-15; C=0 and D=-1E-28

I think the solution you want is 4.641564668612784e-10
The other two solutions are complex.

Answer 5

((square root ((5.37 x 10^-10) x (0.0000135) + [Z])) x [Z]) = 1 x 10^-14
([Z]√((5.37*10^-10)(0.0000135) + [Z])) = 1 x 10^-14
[Z]^2((5.37*10^-10)(1.35*10^-5) + [Z]) = 1 x 10^-28
[Z]^2((7.2495*10^-15) + [Z]) = 1 x 10^-28
7.2495*10^-15)[Z]^3 + [Z]^2 - 1 x 10^-28 = 0

It appears to be a cubic.

Answer 6

It's the holidays! chill for a few days.