Question

Solve the following quadratic by
completing the square.
y = x^2 - 6x+7

Answer 1

`x= 3+√(2), \quad x= 3 -√(2)`

Explanation:

Given quadratic equation:

`y = x^2 - 6x+7`

To complete the square, begin by adding and subtracting the square of half the coefficient of the term in x:

`\implies y = x^2 - 6x+\left((-6)/(2)\right)^2-\left((-6)/(2)\right)^2+7`

`\implies y = x^2 - 6x+\left(-3\right)^2-\left(-3\right)^2+7`

`\implies y = x^2 - 6x+9-9+7`

Factor the perfect square trinomial:

`\implies y=(x-3)^2-9+7`

`\implies y=(x-3)^2-2`

To solve the quadratic, set it to zero and solve for x:

`\implies (x-3)^2-2=0`

`\implies (x-3)^2-2+2=0+2`

`\implies (x-3)^2=2`

`\implies √((x-3)^2)=√(2)`

`\implies x-3= \pm√(2)`

`\implies x-3+3= 3\pm√(2)`

`\implies x= 3\pm√(2)`

Therefore, the solution to the given quadratic equation is:

`x= 3+√(2), \quad x= 3 -√(2)`

Answer 2

• {3 - √2; 3 + √2}

Explanation:

Given

• Quadratic function y = x² - 6x + 7

Solving by completing the square

• x² - 6x + 7 = 0 Given
• x² - 2*3x + 3² - 3² + 7 = 0 Add, subtract 3²
• (x - 3)² - 9 + 7 = 0 Simplify
• (x - 3)² - 2 = 0 Difference of squares
• (x - 3 + √2)(x - 3 - √2) = 0 Factorize
• x - 3 + √2 = 0 and x - 3 - √2 = 0 Solve each root
• x = 3 - √2 and x = 3 + √2 Answer