Next step, multiply both sides by (.01-x)
.012(.01-x)=.01x+x^2
gather terms, put in standard binomial form, use the quadratic equation, and solve for x.
is the standard binomial x^2+.022x-1.2x10^-4
yes. That is the correct quadratic equation. Now solve for x. Then (H^+) = 0.01 + x and pH = -log(H^+).
yes.
To check if the expression x^2 + 0.022x - 1.2x10^-4 is in standard binomial form, we need to simplify it further. The standard binomial form is ax^2 + bx + c, where a, b, and c are constants.
Looking at the given expression x^2 + 0.022x - 1.2x10^-4, we can see that it is not in standard form yet because there are terms with different powers of x (x^2 and x). We need to combine like terms.
First, let's rewrite 1.2x10^-4 in decimal form:
1.2 x 10^-4 = 1.2 x 0.0001 = 0.00012
Now, let's simplify the expression:
x^2 + 0.022x - 0.00012
After combining like terms, we have x^2 + 0.022x - 0.00012 as the simplified form.
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the simplified expression with the quadratic formula, we get:
a = 1
b = 0.022
c = -0.00012
Substituting these values into the formula:
x = (-0.022 ± √(0.022^2 - 4(1)(-0.00012))) / (2(1))
Simplifying further:
x = (-0.022 ± √(0.000484 + 0.00048)) / 2
= (-0.022 ± √0.000964) / 2
= (-0.022 ± 0.03108) / 2
Now, we have two possible solutions:
x = (-0.022 + 0.03108) / 2 = 0.00908 / 2 = 0.00454 (approx.)
x = (-0.022 - 0.03108) / 2 = -0.05308 / 2 = -0.02654 (approx.)
Therefore, the solutions to the quadratic equation x^2 + 0.022x - 0.00012 are approximately x = 0.00454 and x = -0.02654.