Once you've completed the square how do you solve the equation?

Answer 1

It depends upon the equation that you are trying to solve.

for example if you had x² + y² - 4x - 6y = 3

completing the square (in both x and y) would give you (x - 2)² - 4 + (y - 3)² - 9 = 3

at which point you could collect the numbers together on the right to give you (x - 2)² + (y - 3)² = 16 = 4²

and then "read off" the centre of the circle at (2, 3) and its radius of 4.

However, I suspect that you are asking regarding a quadratic, in which case you have got to something like (x - w)² - r = 0

Just proceed as normal with an equation to solve:

(x - w)² - r = 0

→ (x - w)² = r

→ x - w = ±√r

→ x = w ±√r

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For a quadratic of the form ax² + bx +c this gives:

ax² + bx + c = 0

→ x² +(b/a)x + (c/a) = 0

completing the square:

→ (x + (b/2a))² - (b/2a)² + c/a = 0

→ (x + (b/2a))² = b²/4a² - 4ac/4a²

→ (x + (b/2a))² = (b² - 4ac)/4a²

→ x + (b/2a) = ±√((b² - 4ac)/4a²)

→ x = -(b/2a) ±(√(b² - 4ac))/2a

→ x = (-b ±√(b² - 4ac))/2a

which you may recognise as the formula for solving a quadratic equation.