Question

Sergey is solving 5x2 + 20x – 7 = 0. Which steps could he use to solve the quadratic equation by completing the square? Select three options. 5(x2 + 4x + 4) = –7 + 20 x + 2 = Plus or minus StartRoot StartFraction 27 Over 5 EndFraction EndRoot 5(x2 + 4x) = 7 5(x2 + 4x + 4) = 7 + 20 5(x2 + 4x) = –7

Answer 1

`\boxed{\checkmark}\quad x+2=\pm\sqrt{(27)/(5)}`

`\boxed{\checkmark}\quad 5(x^2+4x)=7`

`\boxed{\checkmark}\quad 5(x^2+4x+4)=7+20`

Explanation:

Sergey is solving the quadratic equation 5x² + 20x - 7 = 0.

To solve the quadratic equation by completing the square, begin by moving the constant to the right side of the equation by adding 7 to both sides:

`\implies 5x^2+20x-7+7=0+7`

`\implies 5x^2 + 20x =7`

The next step is to factor out the leading coefficient from the terms on the left side:

`\implies 5(x^2+4x)=7`

Now we need to make the terms inside the parentheses a perfect square trinomial. To do this, add the square of half the coefficient of the term in x inside the parentheses, and add the distributed value to the right side of the equation:

`\implies 5\left(x^2+4x+\left((4)/(2)\right)^2\right)=7+5\left((4)/(2)\right)^2`

`\implies 5(x^2+4x+4)=7+20`

Factor the perfect square trinomial on the left side and simplify the right side of the equation:

`\implies 5(x+2)^2=27`

Divide both sides of the equation by 5:

`\implies (5(x+2)^2)/(5)=(27)/(5)`

`\implies (x+2)^2=(27)/(5)`

Square root both sides of the equation:

`\implies √((x+2)^2)=\sqrt{(27)/(5)}`

`\implies x+2=\pm\sqrt{(27)/(5)}`

Finally, subtract 2 from both sides:

`\implies x+2-2=-2\pm\sqrt{(27)/(5)}`

`\implies x=-2\pm\sqrt{(27)/(5)}`