Completing the square

Before we start discussing a method for solving quadratic equations, we start with the following statement. We can write a term of the form x2+bxx^2+bx as follows:

x2+bx=(x+b2)2−(b2)2\boxed{x^2+bx = \left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2}

and a term of the form x2−bxx^2-bx as follows:

x2−bx=(x−b2)2−(b2)2\boxed{x^2-bx = \left(x-\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2}

This process is called completing the square. Why are these statements correct? Well, expanding the right side of the equation, we get

(x+b2)2−(b2)2=x2+bx+(b2)2−(b2)2=x2+bx\begin{array}{lll} \left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 &=& x^2+bx+\left( \frac{b}{2} \right)^2-\left( \frac{b}{2} \right)^2\\ &=& x^2+bx \end{array}

and we see that we indeed get the right side. The same is true for the second equation:

(x−b2)2−(b2)2=x2−bx+(b2)2−(b2)2=x2−bx\begin{array}{lll} \left(x-\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 &=& x^2-bx+\left( \frac{b}{2} \right)^2-\left( \frac{b}{2} \right)^2\\ &=& x^2-bx \end{array}
Example 1
  1. x2+9x=(x+4.5)2−4.52=(x+4.5)2−20.25x^2+9x = (x+4.5)^2-4.5^2=(x+4.5)^2-20.25
  2. x2−9x=(x−4.5)2−4.52=(x−4.5)2−20.25x^2-9x = (x-4.5)^2-4.5^2=(x-4.5)^2-20.25

There is also a graphical explanation for completing the square of x2+9xx^2+9x, and the name "completing the square" stems from this method:

Exercise 1

Complete the square in the following expressions:

  1. x2−3xx^2-3x

  2. 2x2+10x2x^2+10x

  3. −5x2+25x-5x^2+25x

  4. x2−xx^2-x

  5. −x2−x-x^2-x

  6. x2x^2

Solution1.x2−3x=(x−1.5)2−1.52=(x−1.5)2−2.252.2x2+10x=2(x2+5x)=2((x+2.5)2−2.52)=2(x+2.5)2−12.53.−5x2+25x=−5(x2−5x)=−5((x−2.5)2−2.52)=−5(x−2.5)2+31.254.x2−x=x2−1x=(x−0.5)2−0.52=(x−0.5)2−0.255.−x2−x=−(x2+1x)=−((x+0.5)2−0.52)=−(x+0.5)2+0.256.x2=x2+0x=(x+0)2−02=x2\begin{array}{llll} 1. & x^2-3x & = & (x-1.5)^2-1.5^2\\ & & = & (x-1.5)^2-2.25\\ 2. & 2x^2+10x & = & 2(x^2+5x)\\ & & = & 2((x+2.5)^2-2.5^2)\\ & & = & 2(x+2.5)^2-12.5\\ 3. & -5x^2+25x & = & -5(x^2-5x)\\ & & = & -5((x-2.5)^2-2.5^2)\\ & & = & -5(x-2.5)^2+31.25\\ 4. & x^2-x & = & x^2-1x\\ & & = & (x-0.5)^2-0.5^2\\ & & = & (x-0.5)^2-0.25\\ 5. & -x^2-x & = & -(x^2+1x)\\ & & = & -((x+0.5)^2-0.5^2)\\ & & = & -(x+0.5)^2+0.25\\ 6. & x^2 & = & x^2+0x\\ & & = & (x+0)^2-0^2\\ & & = & x^2\\ \end{array}