Completing the square for quadratic polynomials
Process for completing the square
The zeroes of a single-variable polynomial are the values of that variable at which the polynomial is equal to ???0???.
Completing the square is a method we can use to find the zeroes of a quadratic polynomial.
Another way to say this is that completing the square is a method we can use to solve the corresponding quadratic equation (the equation that has the quadratic polynomial on one side and ???0??? on the other side).
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The solutions of any polynomial equation are called the roots of that equation. So the zeroes of a quadratic polynomial are numerically equal to the roots of the corresponding quadratic equation.
Completing the square is a useful method when it’s not possible to solve for the roots by factoring, because completing the square creates a trinomial that we can factor as the square of a binomial.
The formal way to write a quadratic polynomial is ???ax^2+bx+c???, where ???a??? is the coefficient of the ???x^2??? term, ???b??? is the coefficient of the ???x??? term, and ???c??? is the constant term.
These are the steps we’ll follow every time we want to complete the square in order to find the roots of a quadratic equation ???ax^2+bx+c=0???.
Before we go through the steps, however, we’ll first divide both sides of the equation by ???a??? (if ???a\ne1???), because it will be easier to solve the equation if the coefficient of the ???x^2??? term is ???1???. If we have to do that division, we won’t define ???b??? and ???c??? until after we do it. That is, ???b??? will be the coefficient of the new ???x??? term, and ???c??? will be the new constant term. So we’ll actually be starting with an equation of the form ???x^2+bx+c=0???.
1. Move the constant term to the right side of the equation by subtracting ???c??? from both sides.
???x^2+bx+c-c=0-c???
???x^2+bx=-c???
2. Find ???(b/2)^2???. Take the coefficient of the ???x??? term, divide it by ???2???, and then square the result.
???x^2+bx+\left(\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2???
3. Factor the left side, which is now
???x^2+bx+\left(\frac{b}{2}\right)^2???
To factor this quadratic polynomial, we have to find a pair of factors of ???(b/2)^2??? whose sum is ???b???. The only pair of factors with this property is ???b/2??? and ???b/2???. So the left side of the equation becomes
???\left(x+\frac{b}{2}\right)\left(x+\frac{b}{2}\right)???
Notice that the trinomial
???x^2+bx+\left(\frac{b}{2}\right)^2???
factors as the square of the binomial ???x+(b/2)???:
???x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)\left(x+\frac{b}{2}\right)=\left(x+\frac{b}{2}\right)^2???
So the equation we have to solve is
???\left(x+\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2???
4. Square root both sides of the equation. Remember that the right side will now include a ???\pm??? sign.
???\sqrt{\left(x+\frac{b}{2}\right)^2}=\sqrt{-c+\left(\frac{b}{2}\right)^2}???
???x+\frac{b}{2}=\pm\sqrt{-c+\left(\frac{b}{2}\right)^2}???
5. Solve for ???x??? to get the roots of the original quadratic equation, by subtracting ???b/2??? from both sides.
???x=-\frac{b}{2}\pm\sqrt{-c+\left(\frac{b}{2}\right)^2}???
How to complete the square on a quadratic polynomial
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Find the roots of the quadratic by completing the square
Example
Solve for ???x??? by completing the square.
???x^2+6x+4=0???
There is no pair of factors of ???4??? whose sum is ???6???, so we’ll need to solve by completing the square. Start by moving the constant constant term, ???4???, to the right side of the equation by subtracting ???4??? from both sides.
???x^2+6x+4-4=0-4???
???x^2+6x=-4???
Find ???(b/2)^2???. In this case, ???b=6???.
???\left(\frac{b}{2}\right)^2=\left(\frac{6}{2}\right)^2=3^2=9???
Add ???9??? to both sides of the equation.
???x^2+6x+9=-4+9???
???x^2+6x+9=5???
Factor ???x^2+6x+9???, and notice that it factors as the square of a binomial, ???(x+3)^2???, so our equation becomes
???(x+3)^2=5???
Take the square root of each side of the equation.
???\sqrt{(x+3)^2}=\sqrt{5}???
Since ???x+3??? could be either positive or negative (because ???(x+3)^2??? is equal to ???5???, which is positive), we get
???x+3=\pm\sqrt{5}???
Solve for ???x??? by subtracting ???3??? from both sides. To avoid confusion, put the ???-3??? in front of the ???\pm\sqrt5???.
???x+3-3=-3\pm\sqrt{5}???
???x=-3\pm\sqrt{5}???
The roots of the original quadratic equation are
???x=-3+\sqrt{5}??? and ???x=-3-\sqrt{5}???
Let’s try another example of completing the square.
Completing the square is a useful method when it’s not possible to solve for the roots by factoring, because completing the square creates a trinomial that we can factor as the square of a binomial.
Example
Solve for the variable by completing the square.
???v^3-5v^2+4v=0???
In this case we could solve by factoring since we can first factor out a ???v??? and then factor the quadratic polynomial that remains.
???v(v^2-5v+4)=0???
???(-1)(-4)=4??? and ???-1+(-4)=-5???, so ???v^2-5v+4??? can be factored as ???(v-1)(v-4)???. Therefore, our original equation can be rewritten as
???v(v-1)(v-4)=0???
One solution is ???v=0???, and we’ll set each of the other two factors (???v-1??? and ???v-4???) equal to ???0??? and solve each of the resulting equations for ???v???.
???v-1=0???
???v-1+1=0+1???
???v=1???
and
???v-4=0???
???v-4+4=0+4???
???v=4???
The solutions are therefore
???v=0???, ???v=1???, and ???v=4???
However, we were not asked to solve by factoring, so let’s look at how this problem can be solved by completing the square. Start by factoring out aso that we’ll have a quadratic polynomial inside parentheses.
???v(v^2-5v+4)=0???
Again, one solution is ???v=0???, so now we’ll find the other solutions, that is, the solutions of the equation
???v^2-5v+4=0???
Now that we have a quadratic polynomial on the left side, we can start by moving the constant term, ???4???, to the right side of the equation by subtracting ???4??? from both sides.
???v^2-5v+4-4=0-4???
???v^2-5v=-4???
Find ???(b/2)^2???. In this case, ???b=-5???.
???\left(\frac{b}{2}\right)^2=\left(\frac{-5}{2}\right)^2=\frac{25}{4}???
Add ???25/4??? to both sides of the equation.
???v^2-5v+\frac{25}{4}=-4+\frac{25}{4}???
???v^2-5v+\frac{25}{4}=\frac{9}{4}???
We see that the trinomial on the left side,
???v^2-5v+\frac{25}{4}???
factors as the square of a binomial, ???[v-(5/2)]^2???, so our equation becomes
???\left(v-\frac{5}{2}\right)^2=\frac{9}{4}???
Take the square root of each side of the equation.
???\sqrt{\left(v-\frac{5}{2}\right)^2}=\sqrt{\frac{9}{4}}???
So we get
???v-\frac{5}{2}=\pm\frac{3}{2}???
Solve for ???v??? by adding ???5/2??? to both sides. To avoid confusion, put the ???5/2??? in front of the ???\pm(3/2)???.
???v-\frac{5}{2}+\frac{5}{2}=\frac{5}{2}\pm\frac{3}{2}???
???v=\frac{5}{2}\pm\frac{3}{2}???
If we solve this both ways (adding and subtracting), we get
???v=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4???
???v=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1???
So the solutions are
???v=0???, ???v=1???, and ???v=4???
