Completing the Square
A polynomial equation with a degree equal to two is known as a quadratic equation. While "quad" means four, "quadratic" means to make a square. A quadratic equation in standard form is represented as:
${ax}^{2}+bx+c=0$
where b and c are real numbers such that a does not equal 0.
Since the degree of the above-written equation is two, it will have two roots, or solutions. The roots of polynomials are the values of x that satisfy the equation. There are a handful of methods you can use to find the roots of a quadratic equation.
One of them is called completing the square. Using this method, you have to convert the given equation into a perfect square.
Steps to completing the square
To solve ${ax}^{2}+bx+c=0$ by completing the square:
- Transform the equation so the constant term, c, is alone on the right side.
- If a, the leading coefficient (the coefficient of the x^{2} term) is not equal to 1, divide both sides by a.
- Add the square of half the coefficient of the x-term, (b/(2a))^{2} to both sides of the equation.
- Factor the left side as the square of a binomial.
- Take the square root of both sides. Remember that ${\left(x+q\right)}^{2}=r$ is equivalent to $x+q=\pm \sqrt{r}$ .
- Solve for x.
The above steps can be implemented as shown below.
Consider the quadratic equation
${ax}^{2}+bx+c=0$ (a not equal to 0)
Get c by itself on the right-hand side
${ax}^{2}+bx=-c$
Dividing both sides by a, we get
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
Add ${\left(\frac{b}{2a}\right)}^{2}$ to both sides
${x}^{2}+\left(\frac{b}{a}\right)x+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$
Factor the left-hand side
${\left(\frac{2ax+b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$
Square root both sides
$\frac{2ax+b}{2a}=\sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}$
This might look overwhelming, but recall that a, b, and c are just numbers, so the whole right side is just some constant.
Example 1
Solve ${x}^{2}-6x-3=0$ by completing the square.
${x}^{2}-6x=3$
${x}^{2}-6x+(-3){}^{2}=3+9$
${(x-3)}^{2}=12$
$x-3=\pm \sqrt{12}$
$x-3=\pm 2\sqrt{3}$
$x=3\pm 2\sqrt{3}$
Example 2
Solve $7{x}^{2}-8x+3=0$ by completing the square.
$7{x}^{2}-8x=-3$
${x}^{2}-\frac{8x}{7}=-\frac{3}{7}$
${x}^{2}-\frac{8x}{7}+{(-\frac{4}{7})}^{2}=-\frac{3}{7}+\frac{16}{49}$
${\left(x-\frac{4}{7}\right)}^{2}=\frac{-5}{49}$
$x-\frac{4}{7}=\frac{\sqrt{5}}{7}i$
$x=\frac{4}{7}\pm \sqrt{\frac{\sqrt{5}}{7}i}$
Alternate ways of writing the above steps
The above steps can also be written by defining 2 more terms d and e.
$d=\frac{b}{2a}$
$e=c-\frac{{b}^{2}}{4a}$
We can then write:
$a(x+d{)}^{2}+e=0$
and solve for x in the following way
${(x+d)}^{2}=-\frac{e}{a}$
$x+d=\pm \sqrt{\frac{-e}{a}}$
$x=-d\pm \sqrt{\frac{-e}{a}}$
Then we can either put in our values for a,b, and c, or d and e to get the answers for x.
$a(x+\frac{b}{2a}{)}^{2}+c-\frac{{b}^{2}}{4a}=0$
Further simplification of this will give you the quadratic formula, which is another method for solving quadratic equations.
Practice questions on completing the square
1. Find the roots of the quadratic equation ${x}^{2}+4x-5=0$ by the method of completing the square.
The given quadratic equation is:
${x}^{2}+4x-5=0$
Comparing the equation with the standard form,
$b=4,c=-5$
${(x+\frac{b}{2})}^{2}=-(c-\frac{{b}^{2}}{4})$
So, ${(x+\frac{4}{2})}^{2}=-(5-\frac{{4}^{2}}{4})$
${(x+2)}^{2}=5+4$
${(x+2)}^{2}=9$
$(x+2)=\pm \sqrt{9}$
$(x+2)=\pm 3$
$x+2=3,x+2=-3$
$x=1,-5$
2. Find the roots of the quadratic equation $3{x}^{2}-5x+2=0$ by completing the square.
The given quadratic equation is:
$3{x}^{2}-5x+2=0$
The given equation is not in the form to which we apply the method of completing the square because the coefficient of ${x}^{2}$ is not 1. To make it 1, we need to divide the whole equation by 3:
${x}^{2}-\frac{5}{3}x+\frac{2}{3}=0$
Comparing with the standard form,
$b=\frac{5}{3},c=\frac{2}{3}$
$c-\frac{{b}^{2}}{4}=\frac{2}{3}-\frac{{(-\frac{5}{3})}^{2}}{4}=\frac{2}{3}-\frac{25}{36}=\frac{24-25}{36}=-\frac{1}{36}$
Substituting these values in the equation ${(x+\frac{b}{2})}^{2}=-(c-\frac{{b}^{2}}{4})$ , we get
${(x-\frac{5}{6})}^{2}=\frac{1}{36}$
$(x-\frac{5}{6})=\pm \sqrt{\frac{1}{36}}$
$(x-\frac{5}{6})=\pm \frac{1}{6}$
$x=1,\frac{2}{3}$
Topics related to the Completing the Square
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Practice tests covering the Completing the Square
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Get help learning about completing the square
Quadratic equations can be one of the more difficult concepts in algebra for students to understand. There are a few different ways to solve them, and completing the square can be especially tricky. If your student is having a hard time learning how to solve quadratic equations by completing the square, getting them help from a professional tutor is an excellent idea. An expert tutor can sit with your student in a 1-on-1 setting and walk them through problems step by step until your student understands the process.
Tutors can discover your student's learning style and use it to present concepts like completing the square in a way that your student understands best. They can then move at the pace that is right for your student. If your student is stuck on a particular concept or step, they can take the time necessary to get over the hump. If another concept or step comes easily to your student, their tutor can breeze through it and move on. This makes tutoring as efficient as it is effective. Your student can ask their tutor questions as soon as they think of them so they don't waste time wondering what to do next or worse, doing some problems the wrong way. With their tutor at their side as they complete homework problems, your student can learn how to do them correctly the first time so bad habits don't take hold and make it more difficult to relearn later.
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