How many ml of a 78% acid solution must be mixed with 12ml of a 10% acid solution to make 30% solution?

Approximately 0.6 ml of 78% acid solution has to be mixed with 12ml of a 10% acid solution to make 30% solution

First, set up a table. Fill in the unknowns with variable "x"

"x" ml of 78% acid solution must be mixed with 12 ml of 10 % acid solution to make 30% acid solution

Let us solve for "x"

The table is attached below

From the table below, we can set up two equations

Sum of values of two salts = Value of mixture

Sum of values of two salts = 0.78x + 1.2

Value of mixture = 0.3(x + 12)

Sum of values of two salts = Value of mixture  

Then, 0.78x + 1.2 = 0.3(x + 12)  

0.78x + 1.2 = 0.3x + 3.6  

0.78x – 0.36x = 3.6 – 1.2  

4.2x = 2.4  

x = 0.5714  ≈ 0.6

Hence, approximately 0.6 ml of 78% acid solution has to be mixed

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.