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What is the solution?

What is the solution?-example-1
User Ndarriulat
by
8.0k points

2 Answers

5 votes

Answer: OPTION A

Explanation:

Apply the Quadratic formula, which is:


x=(-b\±√(b^2-4ac))/(2a)

In this case:


a=1\\b=-6\\c=58

Then, you must susbtitute these values into the quadratic formula, as shown below:


x=(-(-6)\±√((-6)^2-4(1)(58)))/(2*1)


x=(6\±√(-196))/(2)

Keep on mind that
i=√(-1), then you can rewrite it as following:


x=(6\±14i)/(2)\\\\\\x=(2(3\±7i))/(2)\\\\x=3\±7i\\\\x_1=3+7i\\x_2=3-7i

User AbdulAhmad Matin
by
8.6k points
1 vote

Answer:

A. {
3+7i,3-7i}

Explanation:

The given equation is
x^2-6x+58=0

Use the quadratic formula with a=1,b=-6 and c=58

Recall the quadratic formula;


x=(-b\pm√(b^2-4ac) )/(2a)

We substitute the given values to get;


x=(--6\pm √((-6)^2-4(1)(58)) )/(2(1))


x=(6\pm √(36-232) )/(2)


x=(6\pm √(-196) )/(2)

Recall that;


√(-1)=i


x=(6\pm 14i)/(2)


x=3\pm 7i


x=3+7i or
x=3-7i

User Darshan Miskin
by
7.9k points

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