Solving quadratic equations

wolfhound · #1 $Old$ November 9th, 2009, 06:18 AM

If I solve a quadratic equation and I get the values for x as
x=1.788887778 and x=1.11382983
what would they usually round to?

is this allowed?
Thanks

Prof Paulo · #2 $Old$ November 9th, 2009, 06:40 AM

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 $\mathbb{R}$ , of course).

Don't try to put they back to the equation, it doesn't work if the values aren't integers.

wolfhound · #3 $Old$ November 9th, 2009, 07:09 AM

Hi but I need to put them back in to check if they are the correct answer,

could you tell me if I have the correct answers

I had to solve using square root formula

stapel · #4 $Old$ November 9th, 2009, 07:14 AM

Quote:

Originally Posted by wolfhound $View Post$

If I solve a quadratic equation and I get the values for x as x=1.78 and x=1.11

These would be the solutions only if the original quadratic were x^2 - 2.89x + 1.9758, or some multiple thereof. Was it?

Quote:

Originally Posted by wolfhound $View Post$

...so they round to x=1.8 and x=1.1

On what basis had you concluded that the solutions were to be rounded to one decimal place?

Quote:

Originally Posted by wolfhound $View Post$

...and I input the values back in
3(-1.1)^2-2(-1.1)-6=0

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:

$x\, =\, \frac{1\, \pm\, \sqrt{19}}{3}$

To check one of the solutions:

$3\left(\frac{1\, +\, \sqrt{19}}{3}\right)^2\, -\, 2\left(\frac{1\, +\, \sqrt{19}}{3}\right)\, -\, 6$

$3\left(\frac{20\, +\, 2\sqrt{19}}{3}\right)\, -\, 2\left(\frac{1\, +\, \sqrt{19}}{3}\right)\, -\, 6$

$\frac{20\, +\, 2\sqrt{19}}{3}\, -\, \frac{2\, +\, 2\sqrt{19}}{3}\, -\, 6$

$\frac{20}{3}\, +\, \frac{2\sqrt{19}}{3}\, -\, \frac{2}{3}\, -\, \frac{2\sqrt{19}}{3}\, -\, 6$

$\frac{18}{3}\, -\, 6$

...and so forth.

wolfhound · #5 $Old$ November 9th, 2009, 07:20 AM

Hi thanks but
I am only allowed use the -b (square root )b^2-4ac/2a formula or standard factoring!
I got these values from -b formula

wolfhound · #6 $Old$ November 9th, 2009, 07:33 AM

would my answer be correct to the - b formula???

HallsofIvy · #7 $Old$ November 9th, 2009, 07:40 AM

Quote:

Originally Posted by wolfhound $View Post$

Hi thanks but
I am only allowed use the -b (square root )b^2-4ac/2a formula or standard factoring!
I got these values from -b formula

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 $3x^2- 2x- 6= 0$ , since you put x= -1.1 (after saying that x= 1.11 was a solution!) into that.

By the quadratic formula, $x= \frac{2\pm\sqrt{4+ 72}}{6}= \frac{2\pm\sqrt{76}}{6}$ . $x= \frac{2+ \sqrt{76}}{2}$ and $x= \frac{2- \sqrt{76}}{2}$ are the correct roots.

$\sqrt{76}$ 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?

wolfhound · #8 $Old$ November 9th, 2009, 07:56 AM

Hi,
I thought I was supposed to round to 2 digits because 76(two digits) was the largest number
Could you tell me what the correct rounding would be PLEASE

kjchauhan · #9 $Old$ November 9th, 2009, 04:01 PM

the exact rounded values are 1.79 and -1.12.

mr fantastic · #10 $Old$ November 9th, 2009, 07:48 PM

Quote:

Originally Posted by wolfhound $View Post$

Hi,
I thought I was supposed to round to 2 digits because 76(two digits) was the largest number
Could you tell me what the correct rounding would be PLEASE

The correct rounding to use is the rounding that the question tells you to do!! Otherwise, give exact surd values. Until you state the entire question, exactly as it was written and with all instructions, there is nothing more can be said.

HallsofIvy · #11 $Old$ November 10th, 2009, 03:56 AM

Quote:

Originally Posted by wolfhound $View Post$

Hi,
I thought I was supposed to round to 2 digits because 76(two digits) was the largest number
Could you tell me what the correct rounding would be PLEASE

Well, you are wrong. In working with measurements, your accuracy cannot be greater than that of the least accurate figure- the one with the fewest significant digits, not most. But there are no measurements involved here so that is irrelevant.