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The values of x such that \(2x^2 - 6x + 5 = 0\)

are \( m+ni\) and \(m-ni,\) where \(m\) and \(n\) are positive. What is \(m\cdot n?\)

Guest Dec 27, 2017
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 #1
avatar+91479 
+3

 

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-b \over2a} \pm {\sqrt{b^2-4ac} \over 2a}\\~\\ m=\frac{-b}{2a}=\frac{--6}{2*2}=\frac{3}{2}\\ ni={\sqrt{b^2-4ac} \over 2a}={\sqrt{(-6)^2-4*2*5} \over 2*2}={\sqrt{36-40} \over 4}={2i \over 4}={1\over 2}i\\\ mn=\frac{3}{2}\times {1\over 2}=\frac{3}{4}\)

Melody  Dec 27, 2017
 #2
avatar+81063 
+1

In the form  ax^2  +  bx  + c  =  0

 

The sum  of  the  roots will be  = -b/a

And the product of the roots will be  = c/a

 

Sum  of the roots   =   2m

Product of the roots  =  m^2  + n^2

 

So

 

2m  =  6/2   ⇒   m  =  3/2

And

m^2  + n^2  =  5/2    ⇒   (9/4) + n^2  =  5/2  ⇒  n^2  = 10/4 - 9/4  =  1/4  ⇒  n  =  1/2

 

So

 

m * n   =   (3/2)(1/2)  =  3/4

 

 

cool cool cool

CPhill  Dec 28, 2017

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