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Completing the Square Practice Math Problems and Examples

Created by Anna Szczepanek, PhD
Reviewed by Rijk de Wet
Last updated: Jun 05, 2023


Here you can find practice questions for the method of solving quadratic equations by completing the square. This is one of the most important problems in high school math! Go through the examples and problems we offer, and become a master of completing the square.

Once you're done here and want to learn another useful method, go to the quadratic formula calculator!

Solving general quadratic equations by completing the square

In high school algebra, you're bound to encounter the problem of solving general quadratic equations by the "completing the square" method. A general quadratic equation is an equation involving a quadratic polynomial (so a polynomial of degree two):

ax2+bx+c=dax^2 + bx + c = d

where aa, bb, cc and dd are real coefficients. It's clear that such an equation can be simplified to get a=1a=1 as the leading coefficient and to get the constant term at the right side of the equation equal to 00. To do that, let's subtract dd from both sides and then divide by aa:

ax2+bx+c=dax2+bx+(cβˆ’d)=0x2+bax+cβˆ’da=0\begin{split} ax^2 + bx + c &= d \\ ax^2 + bx + (c-d) &= 0 \\ x^2 + \frac{b}{a} x + \frac{c-d}{a} &= 0 \\ \end{split}

Consequently, it suffices to know how to solve a quadratic equation of the form:

x2+bx+c=0x^2+bx+c = 0

To do it, add βˆ’c+b24-c+\frac{b^2}{4} to both sides:

x2+bx+b24=βˆ’c+b24x^2+bx+\frac{b^2}{4} = -c+\frac{b^2}{4}

We can see a perfect square trinomial on the left side:

x2+bx+b24=(x+b2)2x^2+bx+\frac{b^2}{4} = \left(x+\frac{b}{2}\right)^2

πŸ’‘ If you struggle to understand what happened here, go to the perfect square trinomial calculator!

Thus,

(x+b2)2=βˆ’c+b24\left(x+\frac{b}{2}\right)^2 = -c+\frac{b^2}{4}

And so now, if the right side is non-negative, we take the square root of both sides β€” the left side of the equation becoming ∣x+b2∣|x + \frac{b}{2}| β€” and perform standard calculations to solve the equation. If the right side is negative, the equation has no real solution (although it does have complex solutions).

Completing the square practice problems quiz

What term do you have to add to the following expression in order to get a perfect square trinomial?

  1. x2+2xx^2 + 2x
  2. x2βˆ’6xx^2 - 6x
  3. x2+3xx^2 + 3x
  4. x2+6x+6x^2 + 6x +6
  5. x2βˆ’2x+3x^2 - 2x + 3

Answers:

  1. Add 11 to get the trinomial (x+1)2(x+1)^2.
  2. Add 99, trinomial: (xβˆ’3)2(x-3)^2.
  3. Add 94\frac{9}{4}, trinomial: (x+32)2(x+\frac{3}{2})^2.
  4. Add 33, trinomial: (x+3)2(x+3)^2.
  5. Subtract 22, trinomial: (xβˆ’1)2(x-1)^2.

Example questions

Here we solve quadratic equations by completing the square so that you can learn this method with some examples.

Example 1. x2βˆ’x+0.25=1x^2 - x + 0.25 = 1

We can expand the left-hand side as

x2βˆ’x+0.25=(xβˆ’0.5)2∴(xβˆ’0.5)2=1\begin{split} x^2 - x + 0.25 &= (x-0.5)^2 \\ \therefore (x-0.5)^2 &= 1 \\ \end{split}

We can apply the square root to both sides to get ∣xβˆ’0.5∣=1|x - 0.5|=1 and therefore we know xβˆ’0.5=Β±1x - 0.5 = \pm 1. Therefore, x=1.5x = 1.5 or x=βˆ’0.5x = -0.5.

Example 2. 2x2+4x+8=02x^2 + 4x + 8 = 0

Let's divide both sides by 22:

x2+2x=0x^2 + 2x = 0

Now, we examine the expression x2+2xx^2 + 2x. To produce these terms by the short multiplication formula, we can use (x+1)2=x2+2x+1(x + 1)^2 = x^2 + 2x + 1. As you can see, the constant term is 11 while in the equation it is 00 β€” we need to complete the square by adding 11:

x2+2x+1=1(x+1)2=1∣x+1∣=1=1x+1=Β±1∴x=0or x=βˆ’2\begin{split} x^2 + 2x + 1 &= 1 \\ (x +1)^2 &= 1 \\ |x + 1| &= \sqrt{1} = 1 \\ x + 1 &= \pm 1 \\ \therefore \qquad x &= 0 \\ \text{or } x &= -2 \\ \end{split}

Example 3. x2βˆ’8x+20=0x^2 - 8x + 20 = 0

At the left side we have x2βˆ’8xx^2 - 8x, which can be interpreted as a part of the perfect square trinomial x2βˆ’8x+16=(xβˆ’4)2x^2 - 8x + 16 = (x-4)^2. However, we have 2020 in our equation and only 1616 in the perfect square trinomial. So let's subtract 44 from both sides:

x2βˆ’8x+20βˆ’4=βˆ’4x2βˆ’8x+16=βˆ’4(xβˆ’4)2=βˆ’4\begin{split} x^2 - 8x + 20 - 4 &= -4 \\ x^2 - 8x + 16 &= -4 \\ (x - 4)^2 &= -4 \\ \end{split}

Clearly, this equation has no solution in real numbers, because no real number can give βˆ’4-4 when squared. Let's solve this equation in complex numbers (if you're not yet familiar with complex numbers, feel free to skip this part). So, we have

(xβˆ’4)2=(2i)2xβˆ’4=Β±2ix=4Β±2i\begin{split} (x - 4)^2 &= (2i)^2 \\ x-4 &= \pm 2i \\ x &= 4 \pm 2i \\ \end{split}

Practice questions

Solve using the method of completing the square:

  1. x2+8xβˆ’9=0x^2 + 8x - 9 = 0
  2. 3x2+12x=03x^2 + 12x = 0
  3. x2+4x=βˆ’4x^2 + 4x = -4

Answers:

  1. x=βˆ’9x = -9 or x=1x = 1
  2. x=0x = 0 or x=βˆ’4x = -4
  3. x=βˆ’2x = -2
Anna Szczepanek, PhD
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