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21 August, 14:28

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 $m/cdot n?$

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  1. 21 August, 15:02
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    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
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