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 $m/cdot n?$

Question

Ryan Fischer

Mathematics

9 March, 05:52

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 $m/cdot n?$

+3

Answers (1)

Know the Answer?

Kason Allison · Answer 1

2x^2-6x+5=0

solving the quadratic above using the quadratic equation we get:

x=[-b+/-sqrt (b^2-4ac) ]/2a

plugging the values we get:

x=[6+/-sqrt (36-4*5*2) ]/2*2

x=3/2-i/2

or

x=3/2+i/2

thus

m=3/2 and n=1/2

thus

m*n=3/2*1/2=3/4

Answer: 3/4

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 $m/cdot n?$

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 $m/cdot n?$