Not every quadratic equation can be solved easily by factoring or by extraction of roots. If the solutions are not integers, the expression \(x^2+bx+c\) may be very difficult to factor. For example, the equation \(x^2 + 3x - 1 = 0\) cannot be solved easily by either of these methods. In this section we learn a method that will solve any quadratic equation.
Subsection3.4.1Squares of Binomials
We can use extraction of roots to solve equations of the form
where the left side of the equation includes the square of a binomial, or a perfect square. It turns out that we can write any quadratic equation in this form.
Study the following squares of binomials.
Square of binomial \((x+p)^2\)
\(p\)
\(2p\)
\(p^2\)
1. \((x+\alert{5})^2=x^2+10x+25\)
\(\alert{5}\)
\(2(\alert{5})=10\)
\(\alert{5}^2=25\)
2. \((x\alert{{}-{}3})^2=x^2-6x+9\)
\(\alert{-3}\)
\(2(\alert{-3})=-6\)
\((\alert{-3})^2=9\)
3. \((x\alert{{}-{}12})^2=x^2-24x+144\)
\(\alert{-12}\)
\(2(\alert{-12})=-24\)
\((\alert{-12})^2=144\)
In each case, the square of the binomial is a quadratic trinomial,
The coefficient of the linear term, \(2p\text{,}\) is twice the constant in the binomial, and the constant term of the trinomial, \(p^2\text{,}\) is its square.
Checkpoint3.4.1.QuickCheck 1.
What is the linear term of \((x+6)^2\) ?
\(\displaystyle x^2\)
\(\displaystyle 6x\)
\(\displaystyle 12x\)
\(\displaystyle 36\)
Answer.
\(\text{Choice 3}\)
Solution.
\(12x\) is the linear term of \(x^2+12x+36\text{.}\)
Checkpoint3.4.2.QuickCheck 1.
What is the linear term of \(~(x+6)^2~\) ?
We would like to reverse the process and write a quadratic expression as the square of a binomial. For example, what constant term can we add to
\begin{equation*}
x^2 - 16x
\end{equation*}
to produce a perfect square trinomial? Compare the expression to the formula above:
You can check that in the resulting trinomial, the constant term is equal to the square of one-half the coefficient of\(x\text{.}\) In other words, we can find the constant term by taking one-half the coefficient of \(x\) and then squaring the result. Adding a constant term obtained in this way is called completing the square.
Example3.4.3.
Complete the square by adding an appropriate constant; write the result as the square of a binomial.
One-half of \(5\) is \(\dfrac{5}{2}\text{,}\) so the constant term is \(\left(\dfrac{5}{2}\right)^2\text{,}\) or \(\dfrac{25}{4}\text{.}\) We add \(\dfrac{25}{4}\) to obtain
Subsection3.4.2Solving Quadratic Equations by Completing the Square
Now we will use completing the square to solve quadratic equations. First, we will solve equations in which the coefficient of the squared term is 1. Consider the equation
Step 3 The left side of the equation is now the square of a binomial, namely \((x - 3)^2\text{.}\) We write the left side in its square form and simplify the right side, which gives us
\begin{equation*}
(x - 3)^2 =16
\end{equation*}
Step 4 We can now use extraction of roots to find the solutions. Taking square roots of both sides, we get
\begin{align*}
x - 3 \amp =4 \amp\text{or}\amp\amp x - 3 \amp= -4\amp \blert{\text{Solve each equation.}}\\
x \amp =7 \amp\text{or}\amp\amp x \amp= -1
\end{align*}
The solutions are \(7\) and \(-1\text{.}\)
In Step 3, you can check that this equation is equivalent to the original one; if you expand the left side and collect like terms, you will return to the original equation:
The graph of \(y = x^2 - 6x - 7\) is shown at left. Notice that the \(x\)-intercepts of the graph are \(x = 7\) and \(x = -1\text{,}\) and the parabola is symmetric about the vertical line halfway between the intercepts, at \(x = 3\text{.}\)
We can also solve \(x^2 - 6x - 7 = 0\) by factoring instead of completing the square. Of course, we get the same solutions by either method. In the next Example, we solve an equation that cannot be solved by factoring.
Example3.4.9.
Solve \(x^2 - 4x - 3 = 0\) by completing the square.
Solution.
We write the equation with the constant term on the right side.
We write the left side as the square of a binomial, and combine terms on the right side:
\begin{equation*}
(x - 2)^2 =7
\end{equation*}
Finally, we use extraction of roots to obtain
\begin{align*}
x - 2 \amp =\sqrt{7} \amp\text{or}\amp\amp x - 2 \amp= -\sqrt{7}\amp \blert{\text{Solve each equation.}}\\
x \amp =2+\sqrt{7} \amp\text{or}\amp\amp x \amp=2 -\sqrt{7}
\end{align*}
These are the exact values of the solutions. We use a calculator to find decimal approximations for each solution:
Our method for completing the square works only if the coefficient of \(x^2\) is 1. If we want to solve a quadratic equation whose lead coefficient is not 1, we first divide each term of the equation by the lead coefficient.
Example3.4.14.
Solve \(~~2x^2 - 6x - 5 = 0\text{.}\)
Solution.
Because the coefficient of \(x^2\) is \(2\text{,}\) we must divide each term of the equation by \(2\text{.}\)
Finally, extract roots and solve each equation for \(x\text{.}\)
\begin{equation*}
x - \frac{3}{2} =\sqrt{\frac{19}{4}} ~~~\text{ or }~~~ x - \frac{3}{2} = -\sqrt{\frac{19}{4}}
\end{equation*}
The solutions are \(\dfrac{3}{2}+\sqrt{\dfrac{19}{4}}\) and \(\dfrac{3}{2}-\sqrt{\dfrac{19}{4}}\text{.}\)
Using a calculator, we can find decimal approximations for the solutions: \(3.679\) and \(-0.679\text{.}\)
Caution3.4.15.
In Example 3.4.14, it is essential that we first divide each term of the equation by \(2\text{,}\) the coefficient of \(x^2\text{.}\) The following attempt at a solution is incorrect.
You can check that \((2x - 3)^2\) is not equal to \(2x^2 - 6x + 9\text{.}\) We have not written the left side of the equation as a perfect square, so the solutions we obtain by extracting roots will not be correct.
Checkpoint3.4.16.Practice 3.
Follow the steps to solve by completing the square:
The diagonal of a rectangle is 20 inches. The width of the rectangle is 4 inches shorter than its length.
Write a quadratic equation about the length of the rectangle.
Solve your equation to find the dimensions of the rectangle.
Answer.
\(\displaystyle L^2+(L-4)^2=20^2\)
12 in by 16 in
22.
The city park used 136 meters of fence to enclose its rectangular rock garden. The diagonal path across the middle of the garden is 52 meters long. What are the dimensions of the garden?
23.
The sail pictured is a right triangle of base and height \(x\text{.}\) It has a colored stripe along the hypotenuse and a white triangle of base and height \(y\) in the lower corner.
Write an expression for the area of the colored stripe.
Express the area of the stripe in factored form.
The sail is \(7 \dfrac{1}{2}\) feet high and the white triangle is \(4 \dfrac{1}{2}\) feet high. Use your answer to part (b) to calculate mentally the area of the stripe.
Answer.
\(\displaystyle A = \dfrac{1}{2}(x^2-y^2)\)
\(\displaystyle A = \dfrac{1}{2}(x-y)(x+y)\)
18 sq ft
24.
An hors d’oeuvres tray has radius \(x\text{,}\) and the dip container has radius \(y\text{.}\)
Write an expression for the area for the chips (the shaded region).
Express the area in factored form.
The tray has radius \(8 \dfrac{1}{2}\) inches and the space for the dip has radius \(2 \dfrac{1}{2}\) inches. Use your answer to part (b) to calculate mentally the area for the chips. Express your answer as a multiple of \(\pi\text{.}\)
25.
Write an expression for the area of the square.
Express the area as a polynomial.
Divide the square into four pieces whose areas are given by the terms of your answer to part (b).
Answer.
\(\displaystyle A = (x+y)^2\)
\(\displaystyle A = x^2+2xy+y^2\)
26.
Write an expression for the area of the shaded region.
Express the area in factored form.
By making one cut in the shaded region, rearrange the pieces into a rectangle whose area is given by your answer to part (b).
Exercise Group.
For Problems 27–30, solve by completing the square. Your answers will involve \(a,~b,~\) or \(c~\) (or a combination of these).
27.
\(x^2+2x+c=0\)
Answer.
\(-1 \pm \sqrt{1-c}\)
28.
\(x^2+bx-4=0\)
29.
\(x^2+bx+c=0\)
Answer.
\(\dfrac{b}{2} \pm \dfrac{b^2-4c}{4}\)
30.
\(ax^2-4x+9=0\)
Exercise Group.
For Problems 31–36, solve the formula for the indicated variable.