Using quadratic formula to prove that if quadratic equation has any non-real roots, then there must be 2 roots which are conjugates of each other?

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Yaretzi Odom · Accepted Answer

Step 1This only applies when all the coefficients are real numbers.For ax2+bx+c=0 you have x=−b±b2−4ac2aFor one root to be non-real you need the discriminant b2−4ac&amp;lt;0.If you let D2=4ac−b2, then the roots can now be written as x=−b2a±iD2a, which are, by definition (same real part, non-zero imaginary part of opposite sign), conjugate complex numbers.

Aliana Sexton · Accepted Answer

Step 1To pursue your idea of using a &quot;difference of two squares&quot;, we can start by writing the quadratic equation in &quot;vertex form&quot; thus,ax2 + bx + c  =  a⋅(x+b2a)2 + (c − b24a⏟k)=  a⋅(x+b2a)2 − (b24a − c)  =  0  ,with the polynomial coefficients all being real numbers.The equation will have non-real roots if the vertex is &quot;above&quot; the  x-axis  (k&amp;gt;0)  for  a&amp;gt;0 and &quot;below&quot; that axis (k&amp;lt;0) when  a&amp;lt;0  ; this is equivalent to the statement  b2−4ac&amp;lt;0 in regard to the discriminant of the quadratic polynomial.The difference of terms can then be factored asa ⋅ [ (x+b2a)2 − ( b2 − 4ac4a2⏟&amp;lt; 0 )2 ]=  a ⋅[ (x+b2a) + i·4ac − b2a ]⋅[ (x+b2a) − i⋅4ac − b2a ]  =  0  ,from which it follows that the zeroes of the quadratic polynomial form the &quot;complex conjugate pair&quot; −b2a ± i⋅4ac − b2a  .

Using quadratic formula to prove that if quadratic equation has

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