How do you find the zeros, real and imaginary, of y=-x^2+4x-42y=−x2+4x−42 using the quadratic formula?

1 Answer
Jun 7, 2016

The roots are x_1=2-sqrt(38)ix1=2−√38i and x_2=2+sqrt(38)ix2=2+√38i.

Explanation:

Function y=ax^2+bx+cy=ax2+bx+c has two roots (zeros) x_{1,2}=(-b+-sqrt(b^2-4ac))/(2a)x1,2=−b±√b2−4ac2a or one double root if b^2=4acb2=4ac.

In this case a=-1,b=4,c=-42a=−1,b=4,c=−42

So you can blindly plug in those values:

x_{1,2}=(-(4)+-sqrt((4)^2-4(-1)(-42)))/(2(-1))=(-4+-sqrt(16-168))/-2 =(4+-sqrt(-152))/2=(4+-2sqrt(38)i)/2=2+-sqrt(38)ix1,2=−(4)±√(4)2−4(−1)(−42)2(−1)=−4±√16−168−2=4±√−1522=4±2√38i2=2±√38i

The roots are x_1=2-sqrt(38)ix1=2−√38i and x_2=2+sqrt(38)ix2=2+√38i.

sqrt(38)=6.164414...

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