Solve by completing the square: x(squared) - 5x = 30?

(x-b)^2 = x^2 - 2bx + b^2

The coefficient of x is -2b. The square of half of that is b^2.

Hence the rule: to make x^2 - k x into a perfect square, add the square of half the coefficient of x.

In your example you add (-5/2)^2 to both sides and the left side becomes the perfect square (x-5/2)^2. Now you need to finish.

Note: it matters not whether you add the square of (-5/2) or add the square of (5/2) because (-5/2)^2 = (5/2)^2.

Note2: you do NOT add the square of (5x/2) nor of (-5x/2). There is no x in the thing you square and add.

As you've already mentioned, the answer is the same.

However, I've always been taught and still advocate this rule. Take the coefficient of x. Therefore, if there is a negative sign, I always write that out as well.

x^2 - 5x - 30 = 0
x^2 - 5x + [(-5)/2]^2 - [(-5)/2]^2 - 30 = 0
(x - (5/2))^2 - (25/4) - 120/4 = 0
(x-(5/2))^2 - 145/4 = 0
x - [5 +/- sqrt(145)]/2 = 0

x²-5x-30=0 x=(5±√25 +120)/2 = 2.5±(√145)/2

x² - 5x = 30

Add the square of half the coefficient of x to both sides

x² - 5x + 25/4 = 30 + 25/4

Factor and simplify

(x - 5/2)² = 145/4

Take square roots

x - 5/2 = ±sqrt(145)/2

x = (5 ± sqrt(145))/2

The reason we can add either the square of 5/2 or -5/2 is that a² = (-a)² for any a.

NEITHER (5x/2) NOR (-5x/2) is squared.
You should be squaring (-5/2).
I agree the minus sign should be included,
even though it won't affect the result of the squaring:
(x - 5/2)^2 = x^2 - 5x + 25/4 =>
(x - 5/2)^2 = 30 + 25/4 =>
x - 5/2 = sqrt(36.25) or (1/2)*sqrt(145).
x = 5/2 +/- (1/2)*sqrt(145).

x**2 is x squared
for ax**2+bx+c = 0,
always x=[-b+sqrt(D)]/2a, x=[-b-sqrt(D)]/2a
where D=b**2-4ac

x^2 - 5x = 30

Divide 5 by 2 to get 5/2

Now square 5/2 to get 25/4

Add and subtract 25/4 to the left hand side of the equation

x^2 - 5x + 25/4 - 25/4 = 30

The first three terms of this equation is a perfect square, i. e

( x - 5/2)^2 - 25/4 = 30

(x - 5/2)^2 = 25/4 + 30

(x - 5/2)^2 = 145/4

x = 5/2 +/- sqrt(145/4)

= 5/2 +/- [sqrt(145)]/2

x² - 5x = 30


I's always the same procedure: you factorize the coefficent of x², here nothing to do, because it's 1.

Then try to make appears one of the 2 identities: (a² + 2ab + b²) or (a² - 2ab + b²).


x² - 5x = 30 → there is the sign "-" → so try with: a² - 2ab + b²

x² - 2.[x * (5/2)] = 30 → you have: a² - 2ab → but b² is missing, so you add 0

x² - 2.[x * (5/2)] + 0 = 30 → you can say that: 0 = + (5/2)² - (5/2)²

x² - 2.[x * (5/2)] + (5/2)² - (5/2)² = 30

{ x² - 2.[x * (5/2)] + (5/2)² } - (5/2)² = 30 → you can see { a² - 2ab + b² } = {a - b}²

{x - (5/2)}² - (5/2)² = 30

{x - (5/2)}² = (5/2)² + 30

{x - (5/2)}² = (25/4) + (120/4)

{x - (5/2)}² = 125/4

{x - (5/2)}² = [± (√125)/2]²

x - (5/2) = ± (√125)/2

x - (5/2) = ± (5√5)/2

x = (5/2) ± [(5√5)/2]

x = (5 ± 5√5)/2

First case: x = (5 + 5√5)/2

Second case: x = (5 - 5√5)/2

x^2 - 5x = 30

x^2 - 2 * x * (5/2) = 30

x^2 - 2x(5/2) + (5/2)^2 - (5/2)^2 = 30

(x - (5/2))^2 - (5/2)^2 = 30

(x - (5/2))^2 = 145/4

x - (5/2) = ± √(145/4)

x = (1/2)(5 ± √145)

When you have x^2 + ax, a greater than zero,
it will be in the form of (x + a/2)^2 - (a/2)^2

When you have x^2 - ax, it will be (x - a/2)^2 - (a/2)