How do you solve the quadratic with complex numbers given #-7x^2+12x=10#?

1 Answer
Oct 1, 2016

Answer:

The two solutions are

#x_1 = 6/7 + i sqrt(34)/7#

#x_2 = 6/7 - i sqrt(34)/7#

Explanation:

First of all, let's bring everything to one side of the equation. I usually like to have the #x^2# coefficient positive, so I'll move everything to the right side, obtaining

#0=7x^2-12x+10#

which of course is the same equation as

#7x^2-12x+10=0#

Now, let's set the usual formula for solving a quadratic equation:

#x_{1,2} = \frac{-b\pm\sqrt{\Delta}}{2a}#

where #\Delta = b^2-4ac#

Let's compute #\Delta# serparately:

#b^2-4ac = (-12)^2-4*7*10 = 144-280 = -136#

To compute the square root of #-136#, we need complex numbers. You can observe that

#sqrt(-136) = sqrt(-1)*sqrt(136) = sqrt(-1)*sqrt(4*34) = sqrt(-1)*sqrt(4)*sqrt(34) = 2isqrt(34)#

So, the solving formula becomes

#x_{1,2} = \frac{12\pm2isqrt(34)}{14} = \frac{6\pmisqrt(34)}{7}#