Explainer: Solving Quadratics by Completing the Square

In this explainer, we will learn how to solve quadratics by completing the square.

We can solve equations in a number of different ways, including factoring, using the quadratic formula, or completing the square. While factoring can be an efficient way to solve an equation, there are many quadratic equations that cannot be solved by factoring. The method of completing the square that we will cover in this explainer allows us to solve any quadratic equation. Completing the square can also give us useful information about the shape of the graph of a quadratic function.

To begin, let us consider the equation π‘₯=25.

If we wished to solve this for π‘₯, we would take the square root of both sides of the equation. We could solve as π‘₯=√25, so π‘₯=5. However, since βˆ’5Γ—βˆ’5 would also give an answer of 25, we can indicate the positive and negative roots here as π‘₯=±√25.

So, π‘₯=Β±5.

It is important to remember that the use of Β± indicates two solutions, so π‘₯=Β±5 represents both π‘₯=5 and π‘₯=βˆ’5.

Let us look at another example of using square roots to find the solution of an equation.

Example 1: Solving a Simple Quadratic Using Square Roots

Solve the equation (π‘₯βˆ’3)=5.

Answer

To begin solving the equation (π‘₯βˆ’3)=5, we would take the square root of both sides, giving π‘₯βˆ’3=√5.

Remember that when taking square roots we must consider both the positive and the negative values. Here, we can indicate the two solution values by writing π‘₯βˆ’3=±√5.

Adding 3 to both sides, we get the final answer: π‘₯=3±√5π‘₯=3+√5π‘₯=3βˆ’βˆš5.and

We can extend this method to solve a quadratic that includes an π‘₯-term. In the previous example, we had an equation that had a perfect, or complete, square of (π‘₯βˆ’3). It was simple to solve as it had just one π‘₯-term. Let us now look at how we can change a quadratic expression in the form π‘₯+𝑏π‘₯+π‘οŠ¨, where 𝑏 and 𝑐 are constants, into an expression with just one π‘₯-term.

Example 2: Completing the Square of an Expression

Complete the square for the expression π‘₯+6π‘₯+20.

Answer

Let us take a visual approach to this question by using a geometric model for the area. We can begin our representation by showing the π‘₯ part of our expression, π‘₯+6π‘₯+20, as a square with a side length of π‘₯.

We can now add the 6π‘₯ term from our expression to the area model. As we want to make a square, we can do this by splitting 6π‘₯ into two rectangles of area 3π‘₯, each with dimensions 3 by π‘₯ units. We can see below how π‘₯+6π‘₯ can be represented.

We now need to find the area of the missing piece of our square. We can calculate from our model that the area of this piece must be 3Γ—3=9 units.

We now have created a complete square of side length (π‘₯+3). We can see in our diagram that (π‘₯+3)=π‘₯+6π‘₯+9.

However, we have the expression π‘₯+6π‘₯+20, so we need to adjust for the extra units.

Since the constant in our original expression is 20, not 9, we can calculate that the additional units must be 11, since 20βˆ’9=11.

We can see below how we have the square, (π‘₯+3), with 11 left over.

Hence, we have completed the square, giving us (π‘₯+3)+11.

Therefore, π‘₯+6π‘₯+20 can be written as the complete square (π‘₯+3)+11.

A visual method, such as the one in the previous example, can be used effectively for completing the square. However, it is useful to have an algebraic method that we can follow regardless of the complexity of the terms in our expressions. Completing the square means changing a quadratic into the square of a binomial, with a constant term added or subtracted.

We will now examine algebraically how we can use the method of completing the square of a quadratic to help us solve an equation in the form π‘₯+𝑏π‘₯+𝑐=0, where where 𝑏 and 𝑐 are constants.

Complete Squares

A complete square can be expressed in the general form (π‘₯+π‘Ž). We can expand the brackets of a complete square (π‘₯+π‘Ž) as (π‘₯+π‘Ž)=π‘₯+2π‘Žπ‘₯+π‘Ž.

Therefore, to complete the square in an equation including π‘₯+2π‘Žπ‘₯, we can substitute π‘₯+2π‘Žπ‘₯=(π‘₯+π‘Ž)βˆ’π‘Ž, into our original equation.

Let us now look at some examples where we complete the square of an equation in order to solve it.

Example 3: Solving a Quadratic Equation by Completing the Square

By completing the square, solve the equation π‘₯+8π‘₯βˆ’10=0.

Answer

Recall that (π‘₯+π‘Ž)=π‘₯+2π‘Žπ‘₯+π‘ŽοŠ¨οŠ¨οŠ¨. If we compare the expressions π‘₯+8π‘₯βˆ’10 and π‘₯+2π‘₯+π‘Ž, we can see that we would equate the coefficients of the π‘₯-terms to get 2π‘Ž=8; hence, π‘Ž=4. So, in order to create a perfect square of (π‘₯+π‘Ž), we take the coefficient of π‘₯ in our given equation and take half of that value as our π‘Ž value. Therefore, we have (π‘₯+4) as our complete square.

On the other hand, we can expand the brackets of (π‘₯+4): (π‘₯+4)=π‘₯+8π‘₯+16.

Subtracting 16 from both sides, we have (π‘₯+4)βˆ’16=π‘₯+8π‘₯.

So, π‘₯+8π‘₯=(π‘₯+4)βˆ’16.

We can now substitute π‘₯+8π‘₯=(π‘₯+4)βˆ’16 into the original equation, π‘₯+8π‘₯βˆ’10=0, giving us (π‘₯+4)βˆ’16βˆ’10=0.

Collecting the constants, this will give us the completed square form (π‘₯+4)βˆ’26=0.

Next, to solve the equation, we perform the same operation to both sides.

Adding 26 to both sides, (π‘₯+4)=26.

When taking the square root, we must consider both the positive and the negative values, so this gives us π‘₯+4=±√26. We can now subtract 4 from both sides to isolate π‘₯, giving us π‘₯=βˆ’4±√26.

Hence, we have two solutions: π‘₯=βˆ’4+√26 and π‘₯=βˆ’4βˆ’βˆš26.

Example 4: Solving a Quadratic Equation by Completing the Square

Solve the equation π‘₯βˆ’14π‘₯+38=0 by completing the square.

Answer

To begin, we put the equation π‘₯βˆ’14π‘₯+38=0 into completed square form.

Recall that π‘₯+2π‘Žπ‘₯=(π‘₯+π‘Ž)βˆ’π‘ŽοŠ¨οŠ¨οŠ¨, where π‘Ž is half of the coefficient of π‘₯.

We can substitute π‘₯βˆ’14π‘₯=(π‘₯βˆ’7)βˆ’49 into the equation π‘₯βˆ’14π‘₯+38=0, giving us (π‘₯βˆ’7)βˆ’49+38=0.

Collecting the constant terms, we have (π‘₯βˆ’7)βˆ’11=0.

We can now add 11 to both sides, giving us (π‘₯βˆ’7)=11.

When taking the square root, we must consider both the positive and the negative values, so this gives us π‘₯βˆ’7=±√11.

Finally, adding 7 to both sides, we have π‘₯=7±√11.

Therefore, π‘₯=7+√11 or π‘₯=7βˆ’βˆš11.

In order to complete the square of a quadratic, we need to have the π‘₯- and π‘₯-terms on the same side of the equation. Let us look at an example where we need to rearrange an equation before solving using the completing the square method.

Example 5: Solving a Quadratic Equation by Completing the Square

By completing the square, solve the equation π‘₯=π‘₯+34.

Answer

To begin solving this, we first need to rearrange the equation π‘₯=π‘₯+34 so that the π‘₯- and π‘₯-terms are on the same side of the equation.

Subtracting π‘₯ from both sides, we have π‘₯βˆ’π‘₯=34.

Recall that π‘₯+2π‘Žπ‘₯=(π‘₯+π‘Ž)βˆ’π‘ŽοŠ¨οŠ¨οŠ¨, where π‘Ž is half of the coefficient of π‘₯. Here, our coefficient of π‘₯ is βˆ’1; therefore π‘Ž=βˆ’12.

Now we can substitute π‘₯βˆ’π‘₯=ο€Όπ‘₯βˆ’12οˆβˆ’ο€Όβˆ’12 into the equation π‘₯βˆ’π‘₯=34, giving us ο€Όπ‘₯βˆ’12οˆβˆ’ο€Όβˆ’12=34.

Evaluating ο€Όβˆ’12, we have ο€Όπ‘₯βˆ’12οˆβˆ’ο€Ό14=34.

Now we can add 14 to both sides to give us ο€Όπ‘₯βˆ’12=34+14ο€Όπ‘₯βˆ’12=1.

When taking the square root, we must consider both the positive and the negative values, so this gives us π‘₯βˆ’12=±√1.

As √1=1, we have π‘₯βˆ’12=Β±1.

Adding 12 to both sides gives us π‘₯=12Β±1.

Therefore, π‘₯=12+1π‘₯=12βˆ’1,π‘₯=32π‘₯=βˆ’12.oror

When we have a quadratic equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, and π‘Ž>1, we simply transform the equation by dividing all the terms by π‘Ž and continue to complete the square as we saw in the previous example.

Example 6: Solving a Quadratic Equation by Completing the Square

By completing the square, solve the equation 3π‘₯βˆ’2π‘₯=2.

Answer

To begin solving an equation with a coefficient of π‘₯ that is not equal to 1, we transformthe equation by dividing everything by the value of this coefficient.

Therefore, dividing 3π‘₯βˆ’2π‘₯=2 by 3, we have π‘₯βˆ’23π‘₯=23.

Recall that π‘₯+2π‘Žπ‘₯=(π‘₯+π‘Ž)βˆ’π‘ŽοŠ¨οŠ¨οŠ¨, where π‘Ž is half of the coefficient of π‘₯. Here, the coefficient of π‘₯ is βˆ’23 so π‘Ž=βˆ’13.

We can now substitute π‘₯βˆ’23π‘₯=ο€Όπ‘₯βˆ’13οˆβˆ’ο€Όβˆ’13 into our equation, π‘₯βˆ’23π‘₯=23, giving us ο€Όπ‘₯βˆ’13οˆβˆ’ο€Όβˆ’13=23.

Evaluating ο€Όβˆ’13 as 19 gives us ο€Όπ‘₯βˆ’13οˆβˆ’19=23.

We can now add 19 to both sides: ο€Όπ‘₯βˆ’13=23+19.

As 23=69, we have ο€Όπ‘₯βˆ’13=79.

When taking the square root, we must consider both the positive and the negative values, so this gives us π‘₯βˆ’13=Β±ο„ž79.

As √9=3, we can write this as π‘₯βˆ’13=±√73.

We can now add 13 to both sides to isolate π‘₯: π‘₯=13±√73.

Hence, our solution is π‘₯=13+√73 or π‘₯=13βˆ’βˆš73.

We have seen in this explainer how we can complete the square to solve a quadratic equation. We will now look at how we can use this process to derive a general formula for solving quadratic equations.

How to: Derive the Quadratic Formula by Completing the Square

Suppose we have a quadratic equation in the general form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants. We can start our process of solving this equation by dividing all the terms by π‘Ž, which gives us π‘₯+π‘π‘Žπ‘₯+π‘π‘Ž=0.

To begin isolating the π‘₯- and π‘₯-terms, we subtract π‘π‘Ž from both sides of the equation, giving us π‘₯+π‘π‘Žπ‘₯=βˆ’π‘π‘Ž.

On the left-hand side of the equation, we can now follow the process for completing the square.

Recall that we can rewrite an expression 𝑐+2π‘π‘‘οŠ¨, with constants 𝑐 and 𝑑, as 𝑐+2𝑐𝑑=(𝑐+𝑑)βˆ’π‘‘.

So, comparing the left-hand side of our equation, we can equate 𝑐=π‘₯ and 𝑑=𝑏2π‘Ž.

Therefore, we can substitute π‘₯+π‘π‘Žπ‘₯=ο€½π‘₯+𝑏2π‘Žο‰βˆ’ο€½π‘2π‘Žο‰οŠ¨οŠ¨οŠ¨ into our equation, giving us ο€½π‘₯+𝑏2π‘Žο‰βˆ’ο€½π‘2π‘Žο‰=βˆ’π‘π‘Ž.

Expanding 𝑏2π‘Žο‰οŠ¨, we have ο€½π‘₯+𝑏2π‘Žο‰βˆ’π‘4π‘Ž=βˆ’π‘π‘Ž.

Adding 𝑏4π‘ŽοŠ¨οŠ¨ to both sides, gives us ο€½π‘₯+𝑏2π‘Žο‰=𝑏4π‘Žβˆ’π‘π‘Ž.

Using 4π‘ŽοŠ¨ as a common denominator on the right-hand side, we can simplify to give ο€½π‘₯+𝑏2π‘Žο‰=π‘βˆ’4π‘Žπ‘4π‘Ž.

We can now take the square root of both sides, remembering the positive and negative values, giving us π‘₯+𝑏2π‘Ž=Β±ο„žπ‘βˆ’4π‘Žπ‘4π‘Žπ‘₯+𝑏2π‘Ž=Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Next, we subtract 𝑏2π‘Ž from both sides, which gives us π‘₯=βˆ’π‘2π‘ŽΒ±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

As both terms on the right-hand side have the same denominator, we can write this in the simplified form as π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

And so we have derived the quadratic formula, which gives a concise solution for all quadratic equations in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0.

Key Points

  • When a quadratic equation has simple factors, we can solve it by factoring. However, completing the square is a useful alternative method for solving quadratics, as it will give a solution to all quadratics, including those that cannot be factored. It is an ideal method if there is an indication in a problem that the solution will involve decimal or fractional solutions.
  • Completing the square can be useful when graphing a quadratic function. When we have a quadratic equation in the form 𝑦=π‘Ž(π‘₯βˆ’β„Ž)+π‘˜οŠ¨, the turning-point vertex will be at (β„Ž,π‘˜). So we can gain additional information about the shape of the graph of a function by completing the square.

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