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# Complex Roots

+1
44
4

The values of x such that

2x^2 - 6x + 5 = -4

are m + ni and m - ni, where m and n are positive.  What is mn?

Apr 6, 2021

#1
+485
+1

$2x^2-6x+9=0$.

Using the quadratic formula, you get $x=\frac{6+\sqrt{36-72}}{4}$ and $x=\frac{6+\sqrt{36-72}}{4}$. These simplify to $x=\frac{3}{2}+6i$ and $x=\frac{3}{2}-6i$.

So mn=$\frac{3}{2}*6=3*3=\boxed{9}$

Apr 6, 2021
#2
+118470
+1

2x^2   - 6x  + 5 =  -4

2x^2  - 6x  =   -9                 complete  the square on x

2(x^2 - 3x  + 9/4)   = -9  + 9/2

2 ( x - 3/2)^2  =   -18/2  + 9/2

2 ( x - 3/2)^2   =  - 9/2

(x - 3/2)^2  =  -9/4          take both roots

x - 3/2  =   (3/2)i           or       x  - 3/2  =  -(3/2)i

x = (3/2)  + (3/2)i         or        x =  3/2  -(3/2)i

m =  n  = 3/2

mn   =  (3/2)^2  = 9/4

Apr 6, 2021
#3
+485
+1

Ohhh, thank you so much @Cphill, I see where I went wrong, I forgot to divide by 4 when I simplified the square roots, so I got 6 instead of $\frac{6}{4}$. Thanks for helping out!

RiemannIntegralzzz  Apr 6, 2021
#4
+118470
+1

Eh....no big deal....easy to  make  mistakes  on here.......(I do it quite often  !!!!)

CPhill  Apr 6, 2021