Need help asap please. thank you!

Question

1. Solve x² - 6x = -8 by completing the square. Enter each step.



2. Solve 4x² + 12x + 11 = 0 by completing the square. Enter each step.

Answer 1

x² - 6x = -8

We must look at the number in front of the term with x, which is -6. Divide -6 by 2 and then square it.

-6 ÷ 2 = -3, then (-3)² = 9

Now add 9 to both sides. x² - 6x + 9 = -8 + 9

Simplify x² - 6x + 9 = 1

Factor the left side (note you should get the same 2 factors) in this case. (x - 3)(x - 3)= 1 or (x - 3)² = 1

Take the square root of both sides and don't forget the ± sign. (x - 3) = ±1

Now consider +1 and -1 separately. x - 3 = 1 and x - 3 = -1

Add 3 to both sides in each case. x = 1 + 3 and x = -1 + 3

Simplify x = 4 and x = 2

The second problem will follow the exact same process, only we need to subtract the 11 from both sides and divide by 4 first, that way it will "look" the same as the first problem.

4x² + 12x + 11 = 0

4x² + 12x + 11 - 11 = 0 - 11

4x² + 12x = -11

x² + 3x = -11/4 (it is important to divide by 4 because we can't have a number in front of x²)

We must look at the number in front of the term with x, which is 3. Divide 3 by 2 and then square it.

3 ÷ 2 = 3/2, then ( 3/2 )² = 9/4

Now add 9/4 to both sides. x² + 3x + 9/4 = -11/4 + 9/4

Simplify x² + 3x + 9/4 = -2/4

Simplify x² + 3x + 9/4 = -1/2

Factor the left side (note you should get the same 2 factors) in this case

(x + 3/2 )(x + 3/2 )= -1/2 or (x + 3/2)² = -1/2

Take the square root of both sides and don't forget the ± and don't forget square root of a negative number gives an imaginary number i, so we get (x + 3/2) = ±i/√2

Now consider +i/√2 and -i/√2 separately. x + 3/2 = i/√2 and x + 3/2 = -i/√2

Add 3/2 to both sides in each case x = i/√2 + 3/2 and x = -i/√2 + 3/2

Rationalize the denominator and Simplify x = i√2/2 + 3/2 and x = - i√2/2 + 3/2

We have common denominator so combine the fractions x = (i√2 + 3)/2 and x = (-i√2 + 3)/2

I will be happy to clarify any and all of the steps if you would like.

Answer 2

1.

x² - 6x = -8

x² - 6x + 8 = 0

(x² - 6x) + 8 = 0

(x² - 6x + C) + (8 - C) = 0

C = Constant to add to the polynomial (x² - 6x) and subtract from 8 to complete the square

Divide the constant multiplied with the x term by 2, and square the result, to find C.

C = (-6/2)²

C = (-3)²

C = 9

(x² - 6x + 9) + (8 - 9) = 0

(x - 3)² + (-1) = 0

(x - 3)² = 1

There are two solutions for x because we square root both sides, and the square root of 1 is -1 and 1.

x - 3 = 1 AND x - 3 = -1

x = 4 AND x = 2

2.

4x² + 12x + 11 = 0

Divide both sides by 4 for simplification of the x² term.

x² + 3x + (11/4) = 0

(x² + 3x + C) + ( (11/4) - C) = 0

C = Constant to add to the polynomial (x² - 6x) and subtract from 8 to complete the square

Divide the constant multiplied with the x term by 2, and square the result, to find C.

C = (3/2)²

C = 9/4

(x² + 3x + (9/4) ) + ( (11/4) - (9/4) ) = 0

(x + (3/2) )² + (1/2) = 0

(x + (3/2) )² = -1/2

(x + (3/2) )² = (-1) * (1/2)

When calculating the square root of a negative number, the result is an imaginary number i.

x + (3/2) = sqrt(-1) * sqrt(1/2)

There are still two solutions, positive and negative, because of the square root operation.

x + (3/2) = i * sqrt(1/2) AND x + (3/2) = - i * sqrt(1/2)

x = i * sqrt(1/2) - (3/2) AND x = - i * sqrt(1/2) - (3/2)