See also: Quadratic Equation, Quadratic Equation Formula, numbers
Given a quadratic equation of the following form,
$$a{x}^{2}+bx+c=0$$we can find the roots by first converting the above form into the one below:
$$a{(x+p)}^{2}+q=0$$where $p=\frac{b}{2a}$ and $q=c-\frac{{b}^{2}}{4a}$ .
The idea of completing the square comes from the fact that if we have a quadratic equation of the form:
$${(x+m)}^{2}=n$$then it's easy to solve by taking the square root of both sides.
$$\begin{array}{rl}\sqrt{{(x+m)}^{2}}& =\pm \sqrt{n}\\ x+m& =\pm \sqrt{n}\\ x& =-m\pm \sqrt{n}\end{array}$$Follow the steps below to solve a quadratic equation in the general form by completing the square.
Original equation | $a{x}^{2}+bx+c=0$ |
Step 1. Divide the equation by $a$ to get the coefficient of ${x}^{2}$ equals to $1$ | ${x}^{2}+\frac{bx}{a}+\frac{c}{a}=0$ |
Step 2. Move the constant term to the right hand side | ${x}^{2}+\frac{bx}{a}=-\frac{c}{a}$ |
Step 3. Add ${\left(\frac{b}{2a}\right)}^{2}$ to both sides of the equation | ${x}^{2}+\frac{bx}{a}+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$ |
Step 4. We can now re-write the left hand side as a complete square | ${(x+\frac{b}{2a})}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$ |
Step 5. Take the square root of both sides | $\begin{array}{rl}\sqrt{{(x+\frac{b}{2a})}^{2}}& =\pm \sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}\\ x+\frac{b}{2a}& =\pm \sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}\end{array}$ |
Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$ | $x=-\frac{b}{2a}\pm \sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}$ |
Examples:
Find the roots of the quadratic equation
$${x}^{2}-4x+3=0$$Step 1. We can skip this step since in this case $a=1$
Step 2. Move the constant term to the right hand side
$${x}^{2}-4x=-3$$Step 3. Add ${\left(\frac{4}{2}\right)}^{2}$ to both sides of the equation (i.e. add $4$)
$$\begin{array}{rl}{x}^{2}-4x+4& =-3+4\\ {x}^{2}-4x+4& =1\end{array}$$Step 4. Re-write left hand side as a complete square
$${(x-2)}^{2}=1$$Step 5. Take the square root of both sides
$$\begin{array}{rl}\sqrt{{(x-2)}^{2}}& =\pm \sqrt{1}\\ x-2& =\pm 1\end{array}$$Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$
$$\begin{array}{rl}x& =2\pm 1\\ {x}_{1}& =2-1=1\\ {x}_{2}& =2+1=3\end{array}$$Find the roots of the quadratic equation
$${x}^{2}-6x+9=0$$Step 1. We can skip this step since in this case $a=1$
Step 2. Move the constant term to the right hand side
$${x}^{2}-6x=-9$$Step 3. Add ${\left(\frac{6}{2}\right)}^{2}$ to both sides of the equation (i.e. add $9$)
$$\begin{array}{rl}{x}^{2}-6x+9& =-9+9\\ {x}^{2}-6x+9& =0\end{array}$$Step 4. Re-write left hand side as a complete square
$${(x-3)}^{2}=0$$Step 5. Take the square root of both sides
$$\begin{array}{rl}\sqrt{{(x-3)}^{2}}& =\pm \sqrt{0}\\ x-3& =0\end{array}$$Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$
$$x=3$$In this case, we could have skipped steps 2 & 3 too, because the original equation is already a complete square, i.e. we can straight away re-write the original equation as a complete square as shown in step 4.
Find the roots of the quadratic equation
$$2{x}^{2}+2x+5=0$$Step 1. Divide the equation by $2$
$${x}^{2}+x+\frac{5}{2}=0$$Step 2. Move the constant term to the right hand side
$${x}^{2}+x=-\frac{5}{2}$$Step 3. Add ${\left(\frac{1}{2}\right)}^{2}$ to both sides of the equation (i.e. add $\frac{1}{4}$)
$$\begin{array}{rl}{x}^{2}+x+\frac{1}{4}& =-\frac{5}{2}+\frac{1}{4}\\ {x}^{2}+x+\frac{1}{4}& =-\frac{9}{4}\end{array}$$Step 4. Re-write left hand side as a complete square
$${(x+\frac{1}{2})}^{2}=-\frac{9}{4}$$Step 5. Take the square root of both sides
$$\begin{array}{rl}\sqrt{{(x+\frac{1}{2})}^{2}}& =\pm \sqrt{-\frac{9}{4}}\\ x+\frac{1}{2}& =\pm \sqrt{-\frac{9}{4}}\end{array}$$Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$
$$\begin{array}{rl}x& =-\frac{1}{2}\pm \sqrt{-\frac{9}{4}}\\ {x}_{1}& =-\frac{1}{2}-\frac{3i}{2}\\ {x}_{2}& =-\frac{1}{2}+\frac{3i}{2}\end{array}$$This quadratic equation has 2 complex roots, as can be expected because the discriminant ( ${b}^{2}-4ac$ ) is negative.
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See also: Quadratic Equation, Quadratic Equation Formula, numbers