Question

Complete each statement in the steps to solve x2 – 6x – 7 = 0 using the process of completing the square. Isolate the constant by both sides of the equation. Add to both sides of x2 – 6x = 7 to form a perfect square trinomial while keeping the equation balanced. Write the trinomial x2 – 6x + 9 as squared. Use the square root property of equality to get x – 3 = ± . Isolate the variable to get solutions of –1 and 7.
ANSWERS:
Adding 7 to
9
x-3
4

Answer 1

Answers: Using the process of completing the square:

1. Isolate the constant by adding 7 to both sides of the equation:

x^2-6x-7+7=0+7

x^2-6x=7

2. Add 9 to both sides of x2 – 6x = 7 to form a perfect square trinomial while keeping the equation balanced:

x^2-6x+9=7+9

x^2-6x+9=16

3. Write the trinomial x2 – 6x + 9 as squared:

(x-3)^2 = 16

4. Use the square root property of equality to get x – 3 = ±4 .

sqrt[ (x-3)^2 ] = ± sqrt(16)

x-3 = ±4

5. Isolate the variable to get solutions of –1 and 7.

x-3 = ±4

x-3+3 = ±4+3

x = ±4+3

x1=-4+3→x1=-1

x2=+4+3→x2=7

Answer 2

The correct steps to solve the equation are as follows;

Isolate the constant from the equation adding 7 by both sides of the equation.

Adding 9 to both sides of x2 – 6x = 7 to form a perfect square trinomial while keeping the equation balanced.

Use the square root property of equality to get x – 3 = ± 4.

Given that

Equation; $\rm x^2 - 6x - 7 = 0$

According to the question

To solve the given equation follow all the steps given below.

Equation; $\rm x^2 - 6x - 7 = 0$

• Step1; Isolate the constant from the equation adding 7 by both sides of the equation.

        $\rm x^2 - 6x - 7 = 0 \\ \\ \rm x^2 - 6x - 7 +7 = 0+7\\ \\ \rm x^2 - 6x =7$

• Step2; Adding 9 to both sides of x2 – 6x = 7 to form a perfect square trinomial while keeping the equation balanced.

       $\rm x^2-6x=7\\ \\ x^2 -6x+9 = 7+9$

• Step3; Write the trinomial x2 – 6x + 9 as (x-3) squared.

                   $\rm x^2 -6x+9 = 7+9\\\\ (x-3)^2= 16$

• Step4; Use the square root property of equality to get x – 3 = ± 4.

                    $\rm (x-3)^2=16\\ \\ (x-3)^2=4^2\\ \\ x-3 = \pm 4$

• Step5; Isolate the variable to get solutions of –1 and 7.

                    $\rm (x-3) = \pm 4\\ \\ x-3=4, \ x=4+3, \ x=7\\ \\ x-3=-4, \ x=-4+3, \ x=-1$

To know more about the Quadratic equation click the link given below.

https://brainly.com/question/3638962