 # Help Needed on Quadratic Question

I am trying to solve the equation 40x^2-18x+8=0 using the most efficient method possible but I can’t seem to figure it out.

It’s because you can’t solve that specific quadratic equation!
The reason I know this is because I used the discriminant on your equation, and it came back with a negative value, which means that your equation only has complex roots.

Basically, the discriminant is the value b^2 - 4ac.
If this value is >0, then your equation has two real roots and you can solve it.
If this value =0, then your equation has one real root and you can solve it.
If this value <0, then your equation has only complex roots and you cannot solve it.

Since, in your case, b^2 - 4ac = (-18)^2 - 4(40)(8) = -956 < 0, we know that your equation only has complex roots and cannot be solved (in terms of real numbers, at least).

Does that help?

But what would be the answer if I did do it by using decimals? I tried and i don’t know if my answer is right

What exactly do you mean by using decimals? Do you mean using the quadratic formula?

If I apply the quadratic formula to this quadratic equation, this is what happens:
\displaystyle x = \frac{b^2 \pm \sqrt{b^2 - 4ac}}{2a} = \frac{(-18)^2 \pm \sqrt{(-18)^2 - 4(40)(8)}}{2(40)} = \frac{(-18)^2 \pm \sqrt{-956}}{80},

but here we are now stuck and can’t proceed further (unless you know about imaginary numbers, i), because we can’t have a negative number under the square-root symbol. Numbers under the square-root always have to be nonnegative. Hence, this quadratic equation has no solution (in real numbers).

Does that make sense?

Yes that makes more sense, thank you so much for your help! I really appreciate it.

No problem at all!

This topic was automatically closed 24 hours after the last reply. New replies are no longer allowed.