((x)(x))/(3-x) = 1.8x10^-5
how do I solve for x
multiply bout sides by (3-x)
x^2 = 1.8x10^-5 (3-x)
x^2 + 1.8 * 10^-5 x - 5.4 *10^-5 = 0
plain old quadratic now
x^2/(3-x) = 1.8*10^-5
x^2 = 5.4*10^-5 - 1.8*10^-5 x
x^2 + 1.8*10^-5 x - 5.4*10^-5 = 0
I guess the quadratic formula is the best way to handle that. I get
x = -0.00735747, 0.00733947
or, more exactly,
x = (-1.8*10^-5 ±√((1.8*10^-5)^2+21.6*10^-5))/2
= (-0.000018±√(3.24*10^-10 + 21.6*10^-5))/2
= (-0.00018±√.000216000324)/2
= (-0.00018±0.0146969)/2
as above
To solve the equation ((x)(x))/(3-x) = 1.8x10^-5, we can follow these steps:
Step 1: Rewrite the equation to eliminate the denominator:
Multiply both sides of the equation by (3 - x) to get rid of the fraction, giving us:
(x)(x) = 1.8x10^-5 * (3 - x)
Simplifying this step, we have:
x^2 = 1.8x10^-5 * (3 - x)
Step 2: Expand and rearrange the equation:
Distribute 1.8x10^-5 to (3 - x):
x^2 = 5.4x10^-5 - 1.8x10^-5x
Rearrange the equation by moving all terms to one side:
x^2 - 5.4x10^-5 + 1.8x10^-5x = 0
Step 3: Combine like terms:
x^2 + x(1.8x10^-5 - 5.4x10^-5) - 5.4x10^-5 = 0
Simplifying further:
x^2 + 1.8x10^-5x - 3.6x10^-5x - 5.4x10^-5 = 0
x^2 - 1.8x10^-5x - 5.4x10^-5 = 0
Step 4: Solve the quadratic equation:
You can either factor the quadratic equation if it is factorable or use the quadratic formula to solve for x. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation: x^2 - 1.8x10^-5x - 5.4x10^-5 = 0, the coefficients are:
a = 1, b = -1.8x10^-5, and c = -5.4x10^-5
Substituting these values into the quadratic formula and simplifying:
x = (-(-1.8x10^-5) ± √((-1.8x10^-5)^2 - 4(1)(-5.4x10^-5))) / (2(1))
x = (1.8x10^-5 ± √((3.24x10^-10) + 2.16x10^-4)) / 2
x = (1.8x10^-5 ± √(2.16x10^-4 + 3.24x10^-10)) / 2
Now you can calculate the two possible values of x by plugging in the coefficients and simplifying the expression under the square root.
Note: In some cases, the equation may not have real solutions or may have complex solutions.