Hello there!
Answer: ⇒ [tex]x=\frac{7}{2}i,x=2i[/tex]
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Step-by-step explanation:
First you had to substitute.
[tex]2(a+bi)^2-11i(a+bi)-14=0[/tex]
Then expand.
[tex]2(a+bi)^2-11i(a+bi)-14=(2a^2-2b^2+11b-14)+i(-11a+4ab)[/tex]
[tex](2a^2-2b^2+11b-14)+i(-11a+4ab)=0[/tex]
Rewrite on in standard complex form: [tex]0+0i[/tex]
[tex](2a^2-2b^2+11b-14)+i(-11a+4ab)=0+0i[/tex]
Complex numbers can be equal only if their real and imaginary parts are equal and rewrite as systems of equal.
[tex](2a^2-2b^2+11b-14=0, -11a+4ab=0)=(b=\frac{7}{2}, a=0, b=2, a=0)[/tex]
Substitute back x=a+bi.
[tex]x=\frac{7}{2}i, x=2i[/tex]
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Hope this helps!
Thank you for posting your question at here on brainly.
Have a great day!
-Charlie
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