Can someone please help me with this math problem?

Question

I really need to know the steps that you took to arrive at the answer.

Solve the quadratic equation by completing the square.
x^2 + 6x + 34 = 0

Answer 1

To complete the square, we will first move the + 34 to the other side by subtracting it from both sides.

x^2 + 6x + 34 - 34 = 0 - 34

x^2 + 6x = - 34 (Rewrite.)

Next, we take 1/2 of the "b" term (coefficient of x), and square it. Add this value onto both sides of the equation.

x^2 + 6x + 36 = - 34 + 9

x^2 + 6x + 9 = -25 (Simplify the right side.)

(x + 3)^2 = -25

Take the square root of both sides.

x + 3 = 5i or x + 3 = - 5i

Solve each equation.

x = -3 + 5i or x = -3 - 5i

Answer 2

Original equation:
x² + 6x + 34 = 0

I like to get the x² and x terms on one side, so subtract 34 from both sides:
x² + 6x = -34

To complete the square, take the coefficient on the x term (6), divide it in half (3) and square it (9). Add this to both sides:
x² + 6x + 9 = -34 + 9

The left side is now a perfect square, so rewrite it:
(x + 3)² = -25

To solve, take the square root of both sides. Note, this will have two irrational roots obviously.
x + 3 = ±5i

Subtract 3 from both sides and you have your final answer:
x = -3 ± 5i

x = -3 + 5i or x = -3 - 5i

Answer 3

Well your equation is of the form: aX^2 + bX + c = 0
in your case are the parameters as follows:
a = 1 , b = 6, c = 34

You must use the following formula:
X1/2 = (-b +/- root(b^2 - (4ac)))/(2a)

X1/2 means the first and the second answer.
+/- means one time you have to try the plus and the other time the minus.
b^2 - (4ac) is called the discriminant of the equation use letter D for it.

If D = 0 then there is one answer, if D > 0 then there are two answers, if D < 0 then there is no answer for it in the Real Number Set then we must try the Complex numbers.

Let's begin solving it, first we will fill the parameters (a,b,c) in the equation and we will try the plus first and then the minus.

X1 = (-6 + root(6^2 - (4.1.34)))/(2.1) = -3 + 5i
X2 = (-6 - root(6^2 - (4.1.34)))/(2.1) = -3 - 5i

We get the answers in the Complex Set because:
D<0, root(-100) = root(5^2.-1) = root(5^2).root(-1) = 5i
don't forget root(-1) = i

Answer 4

x2+6x+34=0
x2+6x=-34
make LHS a square by adding 9 to both sides
x2+6x+9=-34+9
which is the same as
(x+3)2=-25
now taking square root on both sides:
we get
x+3=-5i (where i=imaginary)

therefore
x=-3+5i

Answer 5

To solve:
x² + 6x + 34 = 0
You will need to use the formula;
x = (-b ±√(b²-4ac)) /(2a)

a = 1 (from 1x² )
b = 6 from +6x
c = 34 (from +34)

x = (-b ±√(b²-4ac)) /(2a)

x = [-6 ±√(36-136)] / 2
x = [-6 ±√-100] / 2
x = [-6 ±√(-1*100)] /2
x = [-6 ±10√(-1)] /2
x = i(-6 ±10)/2
x = -8i or 2i
.... the letter i is used in mathematics to represent the square root of -1 (it is only conventionally possible to find the square root of a POSITIVE number, so the root of -1 is assigned a letter to represent it)
ANS: x = x = -8i or 2i

Answer 6

x^2 + 6x + 34 = 0
x^2 + 6x +9 = -34+9
(x-3)^2 = -25

no real sol'ns