Hi,
The values of x that you find are only approximative. The goal of solve a quadratic equation is to find two, one (equals) or zero (in , of course).
Don't try to put they back to the equation, it doesn't work if the values aren't integers.
Hi,
The values of x that you find are only approximative. The goal of solve a quadratic equation is to find two, one (equals) or zero (in , of course).
Don't try to put they back to the equation, it doesn't work if the values aren't integers.
These would be the solutions only if the original quadratic were x^2 - 2.89x + 1.9758, or some multiple thereof. Was it?
On what basis had you concluded that the solutions were to be rounded to one decimal place?
Since you are not plugging the actual solutions back into the original equation, you cannot check to see if they are correct. At best, you can confirm that they are at least close.
FYI: Non-integral values can be checked. But you'd have to work exactly. For instance, 3x^2 - 2x - 6 = 0 solves as:
To check one of the solutions:
...and so forth.
I don't see what that has to do with anything anyone has said here. And you have not answered a very important question- why did you round your answers to one decimal place?
Apparently, your original equation was , since you put x= -1.1 (after saying that x= 1.11 was a solution!) into that.
By the quadratic formula, . and are the correct roots.
is about 8.72 so x is approximately 1.79 or -1.14. I've reduced those to two decimal places because you gave two decimal places as your first answer. What reason do you have to further reduce to 1.8 and -1.1?