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murzikaleks [220]
2 years ago
5

What are the solutions to the equation

Mathematics
1 answer:
frosja888 [35]2 years ago
3 0

Answer:

C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

Step-by-step explanation:

You have the quadratic function 2x^2-x+1=0 to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions ax^2+bx+c=0 with a\neq 0 the Bhaskara's Formula is:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}

x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}

It usually has two solutions.

Then we have  2x^2-x+1=0  where a=2, b=-1 and c=1. Applying the formula:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}

Observation: \sqrt{-1}=i

x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i

And,

x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i

Then the correct answer is option C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

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