Question

Solve for x to the nearest tenth y completing the square: x^2-5x+7=0​

Answer 1

x = (1/2)(5 ± i√3)

Explanation:

x² - 5x + 7 = 0

subtract 7 from both sides

x² - 5x = -7

Use half the x coefficent, -5/2, as the complete the square term

(x - 5/2)² = -7 + (-5/2)²

(x - 5/2)² = -7 + 25/4

(x - 5/2)² = -3/4

Take the square root of both sides

x - 5/2 = ±(√-3) / 2

x - 5/2 = ±(i√3) / 2

Add 5/2 to both sides

x = 5/2 ± (i√3) / 2

factor out 1/2

x = (1/2)(5 ± i√3)

Answer 2

does not have a solution because √-0.75 ≠ R

Explanation:

x^2 - 5x + (5/2)^2 = -7

x^2 - 5x + 6.25 = -7 + 6.25

(x - 2.5)^2 = -0.75

(x - 2.5) = √-0.75

does not have a solution because √-0.75 ≠ R