quadratic equation question on negative number under the square root

calicejackson

New member
Joined
May 3, 2022
Messages
5
I used an online quadratic equation solver and it seemed to disregard the negative number under the square root sign?
The problem is X^2 +6.8^-4 - 3.4^-5 Under the square root is -.0001355, but it treated it as a positive number and not an imaginary number?
 
You did not give us an equation.

Is the equation [imath]x^2 +6.8^{-4} - 3.4^{-5} = 0[/imath]?

If so, there is no negative under the square root sign because

[math]x^2 +6.8^{-4} - 3.4^{-5} = 0 \implies x^2 = -6.8^{-4} + 3.4^{-5} \implies \\ x = \pm \sqrt{\dfrac{1}{454.35424} - \dfrac{1}{2138.1376}} \approx \sqrt{0.0017332291}[/math]
 
That is the equation I was trying to write. I couldn’t find the exponent key in this format, my apologies. When I do the problem, I keep getting -.0001355? I am using a=2, b=.00068 and c=.000034
When I take b squared minus 4ac I get -.0001355.

There is no square sign in what you stated.
I am having trouble finding the correct math keys to type things in, please bear with me. I am using a=1, b=.00068, and c=.000034 and am referring using these values in the quadratic equation.

You did not give us an equation.

Is the equation [imath]x^2 +6.8^{-4} - 3.4^{-5} = 0[/imath]?

If so, there is no negative under the square root sign because

[math]x^2 +6.8^{-4} - 3.4^{-5} = 0 \implies x^2 = -6.8^{-4} + 3.4^{-5} \implies \\ x = \pm \sqrt{\dfrac{1}{454.35424} - \dfrac{1}{2138.1376}} \approx \sqrt{0.0017332291}[/math]

I see I forgot the x in my question.
I am trying to solve if a=1, b=.00068, and c=.000034. And plugging these values into the quadratic equation
 
I used an online quadratic equation solver and it seemed to disregard the negative number under the square root sign?
The problem is X^2 +6.8^-4 - 3.4^-5 Under the square root is -.0001355, but it treated it as a positive number and not an imaginary number?
[imath]x^2 + 0.00068 x + 0.000034 = 0[/imath]. As you say, the discriminant is -0.0001355. So your online calculator is wrong... the solutions are not real but are complex.

BTW: To write numbers in scientific notation using text you write it as 6.8 x 10^(-4) not 6.8^-4. What you wrote is how you would type it into your calculator.

-Dan
 
Sorry. I am still confused. When you first posted, you put a minus sign before [imath]3.4^{-5}[/imath].

Now you seem to be saying that it is [imath]\text {PLUS } 3.4 * 10^{-5}.[/imath]

Please give the exact equation.

There is no problem if the equation is

[imath]x^2 + 0.000068x - 0.0000034 = 0.[/imath]
 
I used an online quadratic equation solver and it seemed to disregard the negative number under the square root sign?
The problem is X^2 +6.8^-4 - 3.4^-5 Under the square root is -.0001355, but it treated it as a positive number and not an imaginary number?
Assuming you meant the equation is:

X^2 + 6.8*10^(-4) * X - 3.4*10^(-5) = 0

In that case you should not have negative number under radical sign.

You NEED to post an image of the original problem.
 
That is the equation I was trying to write. I couldn’t find the exponent key in this format, my apologies. When I do the problem, I keep getting -.0001355? I am using a=2, b=.00068 and c=.000034
When I take b squared minus 4ac I get -.0001355.


I am having trouble finding the correct math keys to type things in, please bear with me. I am using a=1, b=.00068, and c=.000034 and am referring using these values in the quadratic equation.



I see I forgot the x in my question.
I am trying to solve if a=1, b=.00068, and c=.000034. And plugging these values into the quadratic equation
c is negative not positive. So the discriminant is positive. That's where you have gone wrong.
 
[imath]x^2 + 0.00068 x + 0.000034 = 0[/imath]. As you say, the discriminant is -0.0001355. So your online calculator is wrong... the solutions are not real but are complex.

BTW: To write numbers in scientific notation using text you write it as 6.8 x 10^(-4) not 6.8^-4. What you wrote is how you would type it into your calculator.

-Dan
Thank you so much for your time! I felt like I was going crazy, not that the jury still isn't out on that, but at least one problem is resolved. LOL I used an online quadratic equation solver. I guess that is a lesson learned to beware
 
Top