If one values of x of equation 2x^2-6x+k=0 be 1/2 (a+5i) find values of a and k
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A=
![\frac{3}{2} \frac{3}{2}](https://tex.z-dn.net/?f=+%5Cfrac%7B3%7D%7B2%7D+)
K=
![\frac{109}{8} \frac{109}{8}](https://tex.z-dn.net/?f=+%5Cfrac%7B109%7D%7B8%7D+)
K=
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Step-by-step explanation:
complex roots always occur in conjugate pair
2x2−6x+k=0
sum=3
product = 2k
one root is 2α+5i
other root is 2α−5i
sum=2α+5i+2α−5i=3
α=3
product = k = (23+5i)(23−5i)
=49−25i2=49+25=17 /2
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