Question

Solve each quadratic equation by completing the square. Give exact answers--no decimals.

Answer 1

Answer to Q1: (20)

The solution of given equation are -3±4√2i.

The given equation have no real solution.

Explanation:

We have given a quadratic equation.

x²+6x+41 = 0

We have to solve above equation by completing the square.

Adding -41 to both sides of above equation, we have

x²+6x+41-41 = -41

x²+6x = -41

Adding (3)² to both sides of above equation, we have

x²+6x+(3)² = -41+(3)²

(x+3)² = -41+9

(x+3)² = -32

Taking square root to both sides of above equation, we have

x+3 = ±√-32

x+3 = ±4√2i where i = √-1

x = -3±4√2i

Hence, the solution of given equation are -3±4√2i.

Hence, equation have no real solution.

Answer to Q2: (23)

The solution of given equation are -2+2i and -2-2i.

The given equation has no real solution.

Explanation:

We have given a quadratic equation.

7x²+28x+56 = 0

We have to solve above equation by completing the square.

Taking 7 common from given equation, we have

7(x²+4x+8) = 0

x²+4x+8 = 0

Adding -8 to both sides of above equation, we have

x²+4x+8-8 = 0-8

x²+4x = -8

Adding (2)² to both sides of above equation, we have

x²+4x+(2)² =-8+(2)²

(x+2)² = -8+4

(x+2)² = -4

Taking square root to both sides of above equation, we have

x+2 = ±√-4

x+2 = ±√4i where i = √-1

x+2 = ±2i

x = -2±2i

Hence, the solution of given equation are -2+2i and -2-2i.

The given equation has no real solution.

Answer 2

`x_1= -3+4(√(2))i\\\\x_2 = -3-4(√(2))i`

Explanation:

In this problem we have the equation of the following quadratic equation and we want to solve it using the method of square completion:

`x ^ 2 + 6x +41 = 0`

The steps are shown below:

For any equation of the form:
`ax ^ 2 + bx + c = 0`

1. If the coefficient a is different from 1, then take a as a common factor.

In this case
`a = 1`. Then we go directly to step 2

2. Take the coefficient b that accompanies the variable x. In this case the coefficient is 6. Then, divide by 2 and the result squared it.

We have:

`(6)/(2) = 2`

`((6)/(2)) ^ 2 = 9`

3. Add the term obtained in the previous step on both sides of equality

`x ^ 2 + 6x + 9 = -41 +9`

4. Factor the resulting expression, and you will get:

`(x + 3) ^ 2 =-32`

Now solve the equation:

Note that the term
`(x + 3) ^ 2` is always > 0 therefore it can not be equal to -32.

The equation has no solution in real numbers.

In the same way we can find the complex roots:

`(x + 3) ^ 2 = -32\\\\(x + 3) = \±√(-32)\\\\x = -3 \± √(-32)\\\\x = -3 \± 4(√(2))i\\\\x_1= -3+4(√(2))i\\\\x_2 = -3-4(√(2))i`