solve quadratic equation by completing the square

Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses the I'll post another (more challenging) example too.This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. The solution set to the first equation is: #.

So the first equation is equivalent toĪnd the last equation above is satisfied exactly when: Solve: #x^2+6x-16=0# (by completing the square)Įach of the following equations is equivalent (has exactly the same solutions) as the lines before it. Solving an equation by completing the square: We write #x^2+6x+9-9# If we group it this way: #(x^2+6x+9)-9# then we have a perfect square minus #9#

That doesn't change the value of #x^2+6x#, but it does change the way it's written. Of course you can's just add a number to an expression without changing the value of the expression, so if we want to keep the same value we'll have to make up for adding #9#. We can figure out what to use for #n# by realizing that the #6x# in the middle need to be #2nx#. To make it complete, we'd need to add #n^2# to the expression. Notice: the sign on the middle term matches the sign in the middle of the binomial on the left AND the last term is positive in both.Īlso notice that if we allow #n# to be negative, we only need to write and think about #(x+n)^2=x^2+2nx+n^2# (The sign in the midde will match the sign of #n#.)Īn expression like #x^2+6x# may be thought of as an 'incomplete' square. The square of an expression of the form #x+n# or #x-n# is: So long as we are happy calculating square roots, we can now solve any quadratic equation.Ĭompleting the square is also useful for getting the equation of a circle, ellipse or other conic section into standard form.Īllowing us to identify this equation as that of a circle with centre #(2, -3)# and radius #5#. We got it into a form #t^2-k^2 = 0# with #t = (2ax+b)# and #k=sqrt(b^2-4ac)#, eliminating the linear term leaving only squared terms. So having started with a quadratic equation in the form: Use the zero product property to solve x = 9 or x = -1.Ĭompleting the square is an example of a Tschirnhaus transformation - the use of a substitution (albeit implicitly) in order to reduce a polynomial equation to simpler form. Of course, those nice, rational answers tell me that the original problem could have been solved by factoring: (x-9)(x+1)=0 Phew, that's a lot of steps! And it definitely takes practice. Step 7: #x=-2+-sqrt(11)# move the constant to the other side. Step 6: #x+2= +-sqrt(11)# be sure to use the #+-# on the square root! Step 5: Take the square root of both sides: #sqrt((x+2)^2)=sqrt(11)# Step 3: Add that number to both sides #x^2+4x +4= 7 + 4#

Step 2: take half of the '4', and square that number. Step 1: #x^2 + 4x = 7# (move the constant to the opposite side) There are even websites that will solve the problems for you: Solverįor completing the square, I look for two things: '1' as the lead coefficient (on the #x^2#, and an even number for the coefficient on the linear term (the x term). I think you should develop some strategies for which method is best in which circumstances. Can every quadratic be solved by using the completing the square method? Yes, it sure seems so.but, I wouldn't want to do it that way every time!!Ĭertainly, every quadratic can be solved by the quadratic formula, but I also wouldn't want to use it every time either.

Solve quadratic equation by completing the square