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Section 6.2 Solving Quadratic Equations

Not every quadratic equation can be solved by factoring or by extraction of roots. For example, the expression \(x^2 + x - 1\) cannot be factored, so the equation \(x^2 + x - 1 = 0\) cannot be solved by factoring. For other equations, factoring may be difficult. In this section we learn two methods that can be used to solve any quadratic equation.

Subsection Quadratic Formula

Instead of completing the square every time we solve a new quadratic equation, we can complete the square on the general quadratic equation,

\begin{equation*} ax^2 + bx + c = 0\text{, }~~~ a \ne 0 \end{equation*}

and obtain a formula for the solutions of any quadratic equation.

The Quadratic Formula.

The solutions of the equation \(ax^2 + bx + c = 0\text{, }~~~ a \ne 0\text{,}\) are

\begin{equation*} \blert{x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \end{equation*}

This formula expresses the solutions of a quadratic equation in terms of its coefficients. (The proof of the formula is considered in the Homework problems.) The symbol \(\pm\text{,}\) read plus or minus, is used to combine the two equations

\begin{equation*} x =\dfrac{-b + \sqrt{b^2 - 4ac}}{2a} ~~~~ \text{and} ~~~~x =\dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \end{equation*}

into a single equation.

To solve a quadratic equation using the quadratic formula, all we have to do is substitute the coefficients \(a\text{,}\) \(b\text{,}\) and \(c\) into the formula.

Solve \(~~2x^2 + 1 = 4x\text{.}\)

Solution

Write the equation in standard form as

\begin{equation*} 2x^2 - 4x + 1 = 0 \end{equation*}

Substitute \(\alert{2}\) for \(a\text{,}\) \(\alert{-4}\) for \(b\text{,}\) and \(\alert{1}\) for \(c\) into the quadratic formula, then simplify.

\begin{equation*} \begin{aligned}[t] x \amp =\frac{-(\alert{-4})\pm \sqrt{(\alert{-4})^2 - 4(\alert{2})(\alert{1})}}{2(\alert{2})} \\ \amp = \frac{4 \pm \sqrt{8}}{4} \end{aligned} \end{equation*}

Using a calculator, we find that the solutions are approximately \(1.707\) and \(0.293\text{.}\)

We can also verify that the \(x\)-intercepts of the graph of \(y = 2x^2 - 4x + 1\) are approximately \(1.707\) and \(0.293\text{,}\) as shown below.

parabola with irrational x-intercepts

Use the quadratic formula to solve \(~~x^2 - 3x = 1\text{.}\)

Answer

\(x=\dfrac{3\pm \sqrt{13}}{2} \)

Subsection Applications

We have now seen four different algebraic methods for solving quadratic equations:

  1. Factoring
  2. Extraction of roots
  3. Completing the square
  4. Quadratic formula

Factoring and extraction of roots are relatively fast and simple, but they do not work on all quadratic equations. The quadratic formula will work on any quadratic equation.

The owners of a day-care center plan to enclose a divided play area against the back wall of their building, as shown below. They have \(300\) feet of picket fence and would like the total area of the playground to be \(6000\) square feet. Can they enclose the playground with the fence they have, and if so, what should the dimensions of the playground be?

two fenced rectangles against wall
Solution

Suppose the width of the play area is \(x\) feet. Because there are three sections of fence along the width of the play area, that leaves \(300 - 3x\) feet of fence for its length. The area of the play area should be \(6000\) square feet, so we have the equation

\begin{equation*} x(300 - 3x) = 6000 \end{equation*}

This is a quadratic equation. In standard form,

\begin{equation*} \begin{aligned}[t] 3x^2 - 300x + 6000 \amp= 0\amp\amp \blert{\text{Divide each term by 3.}}\\ x^2 - 100x + 2000 \amp= 0 \end{aligned} \end{equation*}

The left side cannot be factored, so we use the quadratic formula with \(a = \alert{1}\text{,}\) \(b = \alert{-100}\text{,}\) and \(c = \alert{2000}\text{.}\)

\begin{equation*} \begin{aligned}[t] x \amp=\frac{-(\alert{-100}) \pm\sqrt{(\alert{-100})^2 - 4(\alert{1})(\alert{2000})}}{2(\alert{1})}\\ \amp= \frac{100 \pm\sqrt{2000}}{2}\approx \frac{100 \pm 44.7}{2} \end{aligned} \end{equation*}

Simplifying the last fraction, we find that \(x \approx 72.35\) or \(x\approx 27.65\text{.}\) Both values give solutions to the problem.

  • If the width of the play area is \(72.35\) feet, then the length is \(300 - 3(72.35)\text{,}\) or \(82.95\) feet.
  • If the width is \(27.65\) feet, the length is \(300 - 3(27.65)\text{,}\) or \(217.05\) feet.

In Investigation 6.1, we considered the height of a baseball, given by the equation

\begin{equation*} h = -16t^2 + 64t + 4 \end{equation*}

Find two times when the ball is at a height of \(20\) feet. Give your answers to two decimal places.

Answer

\(0.27\) sec, \(3.73\) sec

Sometimes it is useful to solve a quadratic equation for one variable in terms of the others.

Solve \(~~x^2 - xy + y = 2~~\) for \(x\) in terms of \(y\text{.}\)

Solution

We first write the equation in standard form as a quadratic equation in the variable \(x\text{.}\)

\begin{equation*} x^2 - yx + (y - 2) = 0 \end{equation*}

Expressions in \(y\) are treated as constants with respect to \(x\text{,}\) so that \(a = \alert{1}\text{,}\) \(b = \alert{-y}\text{,}\) and \(c = \alert{y - 2}\text{.}\) Substitute these expressions into the quadratic formula.

\begin{equation*} \begin{aligned}[t] x \amp=\frac{-(\alert{-y}) \pm\sqrt{(\alert{-y})^2 - 4(\alert{1})(\alert{y - 2})}}{2(\alert{1})}\\ \amp = \frac{y \pm\sqrt{y^2 - 4y + 8}}{2} \end{aligned} \end{equation*}

Solve \(~~2x^2 + kx + k^2 = 1~~\) for \(x\) in terms of \(k\text{.}\)

Answer

\(x=\dfrac{-k\pm\sqrt{8-7k^2}}{4} \)

Subsection Introduction to complex numbers

You know that not all quadratic equations have real solutions.

For example, the graph of

\begin{equation*} f(x) = x^2 - 2x + 2 \end{equation*}

has no \(x\)-intercepts (as shown at right), and the equation

\begin{equation*} x^2 - 2x + 2 = 0 \end{equation*}

has no real solutions.

parabola without x-intercepts

We can still use completing the square or the quadratic formula to solve the equation.

Solve the equation \(x^2 - 2x + 2 = 0\) by using the quadratic formula.

Solution

We substitute \(a = 1\text{,}\) \(b =-2\text{,}\) and \(c= 2\) into the quadratic formula to get

\begin{equation*} x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(2)}}{2(1)}=\frac{2\pm\sqrt{-4}}{2} \end{equation*}

Because \(\sqrt{-4}\) is not a real number, the equation \(x^2 - 2x + 2 = 0\) has no real solutions.

Solve the equation \(x^2 - 6x + 13 = 0\) by using the quadratic formula.

Answer

\(x=\dfrac{6\pm\sqrt{-16} }{2} \)

Subsubsection Imaginary Numbers

Although square roots of negative numbers such as \(\sqrt{-4}\) are not real numbers, they occur often in mathematics and its applications.

Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. René Descartes gave them the name imaginary numbers, which reflected the mistrust with which mathematicians regarded them at the time. Today, however, such numbers are well understood and used routinely by scientists and engineers.

We begin by defining a new number, \(i\text{,}\) whose square is \(-1\text{.}\)

Imaginary Unit.

We define the imaginary unit \(i\) by

\begin{equation*} i^2=-1 ~~~\text{ or }~~~ i=\sqrt{-1} \end{equation*}
Caution 6.26.

The letter \(i\) used in this way is not a variable; it is the name of a specific number and hence is a constant.

The square root of any negative number can be written as the product of a real number and \(i\text{.}\) For example,

\begin{align*} \sqrt{-4}\amp=\sqrt{-1\cdot 4}\\ \amp= \sqrt{-1}\sqrt{4}=i\cdot2 \end{align*}

or \(\sqrt{-4}=2i\text{.}\) Any number that is the product of \(i\) and a real number is called an imaginary number.

Imaginary Numbers.

For \(a\gt 0\text{,}\)

\begin{equation*} \sqrt{-a}=\sqrt{-1}\cdot\sqrt{a}=i\sqrt{a} \end{equation*}

Examples of imaginary numbers are

\begin{equation*} 3i\text{, }~~~\frac{7}{8}i\text{, } ~~~-38i\text{, }~~~\text{ and }~~~ i\sqrt{5} \end{equation*}

Write each radical as an imaginary number.

  1. \(\displaystyle \sqrt{-25}\)
  2. \(\displaystyle 2\sqrt{-3}\)
Solution
  1. \(\displaystyle \begin{aligned}[t]\\ \sqrt{-25}\amp=\sqrt{-1}\sqrt{25}\\ \amp=i\sqrt{25}=5i \end{aligned}\)
  2. \(\displaystyle \begin{aligned}[t]\\ 2\sqrt{-3}\amp=2\sqrt{-1}\sqrt{3}\\ \amp=2i\sqrt{3} \end{aligned}\)

Write each radical as an imaginary number.

  1. \(\displaystyle \sqrt{-18}\)
  2. \(\displaystyle -6\sqrt{-5}\)
Answer

  1. \(3i\sqrt{2} \)

  2. \(-6i\sqrt{5} \)

Note 6.29.

Every negative real number has two imaginary square roots, \(i\sqrt{a}\) and \(-i\sqrt{a}\text{,}\) because

\begin{equation*} \left(i\sqrt{a}\right)^2=i^2(\sqrt{a})^2=-a \end{equation*}

and

\begin{equation*} \left(-i\sqrt{a}\right)^2=(-i)^2(\sqrt{a})^2=-a \end{equation*}

For example, the two square roots of \(-9\) are \(3i\) and \(-3i\text{.}\)

Subsubsection Complex Numbers

Consider the quadratic equation

\begin{equation*} x^2 - 2x + 5 = 0 \end{equation*}

Using the quadratic formula to solve the equation, we find

\begin{equation*} x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2}=\frac{2\pm\sqrt{-16}}{2} \end{equation*}

If we now replace \(\sqrt{-16}\) with \(4i\text{,}\) we have

\begin{equation*} x=\frac{2\pm4i}{2}=1\pm2i \end{equation*}

The two solutions are \(1 + 2i\) and \(1 - 2i\text{.}\) These numbers are examples of complex numbers.

Complex Numbers.

A complex number can be written in the form \(a+bi\text{,}\) where \(a\) and \(b\) are real numbers.

Examples of complex numbers are

\begin{equation*} 3-5i\text{, }~~~2+\sqrt{7}i\text{, }~~~\frac{4-i}{3}\text{, }~~~6i\text{, }~~~\text{ and } -9 \end{equation*}

In a complex number \(a+bi\text{,}\) \(a\) is called the real part, and \(b\) is called the imaginary part. All real numbers are also complex numbers (with the imaginary part equal to zero). A complex number whose real part equals zero is called a pure imaginary number.

Write the solutions to Example 6.24, \(\dfrac{2\pm\sqrt{-4}}{2}\text{,}\) as complex numbers.

Solution

Because \(\sqrt{-4}=\sqrt{-1}\sqrt{4}=2i\text{,}\) we have \(\dfrac{2\pm\sqrt{-4}}{2}=\dfrac{2\pm2i}{2}\text{,}\) or \(1\pm i\text{.}\) The solutions are \(1+i\) and \(1-i\text{.}\)

Use extraction of roots to solve \((2x + 1)^2 + 9 = 0\text{.}\) Write your answers as complex numbers.

Answer

\(x=\dfrac{-1}{2}\pm\dfrac{3}{2}i \)

Subsection Arithmetic of Complex Numbers

All the properties of real numbers listed in Algebra Skills Refresher Section A.13 are also true of complex numbers. We can carry out arithmetic operations with complex numbers.

We add and subtract complex numbers by combining their real and imaginary parts separately. For example,

\begin{align*} (4 + 5i ) + (2 - 3i ) \amp= (4 + 2) + (5 - 3)i\\ \amp= 6 + 2i \end{align*}
Sums and Differences of Complex Numbers.
\begin{equation*} (a + bi) + (c + di) = (a + c) + (b + d)i \end{equation*}
\begin{equation*} (a + bi) - (c + di) = (a - c) + (b - d)i \end{equation*}

Subtract: \((8 - 6i ) - (5 + 2i )\text{.}\)

Solution

Combine the real and imaginary parts.

\begin{align*} (8 - 6i ) - (5 + 2i ) \amp= (8 - 5) + (-6 - 2)i\\ \amp = 3 + (-8)i = 3 - 8i \end{align*}

Subtract: \((-3 + 2i ) - (-3 - 2i )\text{.}\)

Answer

\(4i\)

Subsection Section Summary

Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.

  • Quadratic trinomial

  • Complete the square

  • Quadratic formula

Subsubsection CONCEPTS

  1. The square of the binomial is a quadratic trinomial,

    \begin{equation*} (x + p)^2 = x^2 + 2px + p^2 \end{equation*}
  2. To Solve a Quadratic Equation by Completing the Square.

      1. Write the equation in standard form.

      2. Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides.

    2. Complete the square on the left side:

      1. Multiply the coefficient of the first-degree term by one-half, then square the result.
      2. Add the value obtained in (a) to both sides of the equation.

    3. Write the left side of the equation as the square of a binomial. Simplify the right side.

    4. Use extraction of roots to finish the solution.

  3. The Quadratic Formula.

    The solutions of the equation \(ax^2 + bx + c = 0\text{, }~~~ a \ne 0\text{,}\) are

    \begin{equation*} \blert{x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \end{equation*}

  4. We have four methods for solving quadratic equations: extracting of roots, factoring, completing the square, and using the quadratic formula. The first two methods are faster, but they don't work on all equations. The last two methods work on any quadratic equation.

Subsubsection STUDY QUESTIONS

  1. Name four algebraic methods for solving a quadratic equation.

  2. Give an example of a quadratic trinomial that is the square of a binomial.

  3. What number must be added to \(x^2 - 26x\) to make it the square of a binomial?

  4. After completing the square, how do we finish solving the quadratic equation?

  5. What is the first step in solving the equation \(2x^2 - 6x = 5\) by completing the square?

Subsubsection SKILLS

Practice each skill in the Homework  problems listed.

  1. Solve quadratic equations by completing the square: #3–24

  2. Solve quadratic equations by using the quadratic formula: #27–36

  3. Solve problems by writing and solving quadratic equations: #37–44

  4. Solve formulas: #45–64

Exercises Homework 6.2

For Problems 1-2, complete the square and write the result as the square of a binomial.

1.

  1. \(x^2+8x\)

  2. \(x^2-7x\)

  3. \(x^2+\dfrac{3}{2}x \)

  4. \(x^2-\dfrac{4}{5}x \)

2.

  1. \(x^2-14x\)

  2. \(x^2+3x\)

  3. \(x^2-\dfrac{5}{2}x \)

  4. \(x^2+\dfrac{2}{3}x \)

For Problems 3-18, solve by completing the square.

3.

\(x^2 - 2x + 1 = 0\)

4.

\(x^2 +4x + 4 = 0\)

5.

\(x^2 +9x + 20 = 0 \)

6.

\(x^2 - x - 20 = 0\)

7.

\(x^2 = 3 - 3x\)

8.

\(x^2 = 5 - 5x\)

9.

\(2x^2 + 4x - 3 = 0\)

10.

\(3x^2 + 12x + 2 = 0\)

11.

\(3x^2 + x = 4\)

12.

\(4x^2 + 6x = 3\)

13.

\(4x^2 - 3 = 2x\)

14.

\(2x^2 - 5 = 3x\)

15.

\(3x^2 - x - 4 = 0\)

16.

\(2x^2 - x - 3 = 0\)

17.

\(5x^2 + 8x = 4\)

18.

\(9x^2 - 12x - 5 = 0\)

For Problems 19-24, solve by completing the square. Your answers will involve \(a\text{,}\) \(b\text{,}\) or \(c\text{.}\)

19.

\(x^2 + 2x + c = 0 \)

20.

\(x^2 - 4x + c = 0 \)

21.

\(x^2 + bx + 1 = 0 \)

22.

\(x^2 + bx - 4 = 0 \)

23.

\(ax^2 + 2x - 4 = 0 \)

24.

\(ax^2 - 4x + 9 = 0 \)

25.

  1. Write an expression for the area of the square in the figure.

  2. Express the area as a polynomial.

  3. Divide the square into four pieces whose areas are given by the terms of your answer to part (b).

square
26.

  1. Write an expression for the area of the shaded region in the figure.

  2. Express the area in factored form.

  3. By making one cut in the shaded region, rearrange the pieces into a rectangle whose area is given by your answer to part (b).

shaded square

For Problems 23-36, solve using the quadratic formula. Round your answers to three decimal places.

27.

\(x^2 - x - 1 = 0\)

28.

\(x^2 + x + 1 = 0\)

29.

\(y^2 + 2y = 5\)

30.

\(y^2 - 4y = -4\)

31.

\(3z^2 = 4.2z + 1.5\)

32.

\(2z^2 = 7.5z - 6.3\)

33.

\(0=x^2- \dfrac{5}{3}x+\dfrac{1}{3} \)

34.

\(0 = -x^2 + \dfrac{5}{2}x-\dfrac{1}{2} \)

35.

\(-5.2z^2 + 176z + 1218 = 0 \)

36.

\(15z^2 - 18z - 2750 = 0 \)

37.

A car traveling at \(s\) miles per hour on a dry road surface requires approximately \(d\) feet to stop, where \(d\) is given by the function

\begin{equation*} d = f(s) = \dfrac{s^2}{24}+\dfrac{s}{2} \end{equation*}

  1. Make a table showing the stopping distance, \(d\text{,}\) for speeds of \(10\text{,}\) \(20\text{,}\) \(\ldots\) , \(100\) miles per hour. (Use the Table feature of your calculator.)

  2. Graph the function for \(d\) in terms of \(s\text{.}\) Use your table values to help you choose appropriate window settings.

  3. Write and solve an equation to answer the question: If a car must be able to stop in \(50\) feet, what is the maximum safe speed it can travel?

38.

A car traveling at \(s\) miles per hour on a wet road surface requires approximately \(d\) feet to stop, where \(d\) is given by the function

\begin{equation*} d = f(s) = \dfrac{s^2}{12}+\dfrac{s}{2} \end{equation*}

  1. Make a table showing the stopping distance, \(d\text{,}\) for speeds of \(10\text{,}\) \(20\text{,}\) \(\ldots\) , \(100\) miles per hour. (Use the Table feature of your calculator.)

  2. Graph the function for \(d\) in terms of \(s\text{.}\) Use your table values to help you choose appropriate window settings.

  3. Insurance investigators at the scene of an accident find skid marks \(100\) feet long leading up to the point of impact. Write and solve an equation to discover how fast the car was traveling when it put on the brakes. Verify your answer on your graph.

39.

A skydiver jumps out of an airplane at \(11,000\) feet. While she is in free-fall, her altitude in feet \(t\) seconds after jumping is given by the function

\begin{equation*} h = f(t) = -16t^2 - 16t + 11,000 \end{equation*}

  1. Make a table of values showing the skydiver's altitude at \(5\)-second intervals after she jumps from the airplane. (Use the Table feature of your calculator.)

  2. Graph the function. Use your table of values to choose appropriate window settings.

  3. If the skydiver must open her parachute at an altitude of \(1000\) feet, how long can she free-fall? Write and solve an equation to find the answer.

  4. If the skydiver drops a marker just before she opens her parachute, how long will it take the marker to hit the ground? (Hint: The marker continues to fall according to the equation given above.)

  5. Find points on your graph that correspond to your answers to parts (c) and (d).

40.

A high diver jumps from the \(10\)-meter springboard. His height in meters above the water \(t\) seconds after leaving the board is given by the function

\begin{equation*} h = f(t) = -4.9t^2 +8t + 10 \end{equation*}

  1. Make a table of values showing the diver's altitude at \(0.25\)-second intervals after he jumps from the airplane. (Use the Table feature of your calculator.)

  2. Graph the function. Use your table of values to choose appropriate window settings.

  3. How long is it before the diver passes the board on the way down?

  4. How long is it before the diver hits the water?

  5. Find points on your graph that correspond to your answers to parts (c) and (d).

41.

A dog trainer has \(100\) meters of chain link fence. She wants to enclose \(250\) square meters in three pens of equal size, as shown in the figure.

three adjacent pens

  1. Let \(l\) and \(w\) represent the length and width, respectively, of the entire area. Write an equation about the amount of chain link fence.

  2. Solve your equation for \(l\) in terms \(w\text{.}\)

  3. Write and solve an equation in \(w\) for the total area enclosed.

  4. Find the dimensions of each pen.

42.

An architect is planning to include a rectangular window topped by a semicircle in his plans for a new house, as shown in the figure. In order to admit enough light, the window should have an area of \(120\) square feet. The architect wants the rectangular portion of the window to be \(2\) feet wider than it is tall.

rectangle surmounted by semicircle

  1. Let \(x\) stand for the horizontal width of the window. Write expressions for the height of the rectangular portion and for the radius of the semicircular portion.

  2. Write an expression for the total area of the window.

  3. Write and solve an equation to find the width and overall height of the window.

43.

When you look down from a height, say a tall building or a mountain peak, your line of sight is tangent to the Earth at the horizon, as shown in the figure.

tangent to earth

  1. Suppose you are standing on top of the Petronas Tower in Kuala Lumpur, \(1483\) feet high. How far can you see on a clear day? (You will need to use the Pythagorean theorem and the fact that the radius of the Earth is \(3960\) miles. Do not forget to convert the height of the Petronas Tower to miles.)

  2. How tall a building should you stand on in order to see \(100\) miles?

44.

  1. If the radius of the Earth is \(6370\) kilometers, how far can you see from an airplane at an altitude of \(10,000\) meters? (Hint: See Problem 43.)

  2. b. How high would the airplane have to be in order for you to see a distance of \(10\) kilometers?

For Problems 45-52, use the quadratic formula to solve each equation for the indicated variable.

45.

\(A = 2w^2 + 4lw ,~~\) for \(w\)

46.

\(A = \pi r^2 + \pi rs ,~~\) for \(r\)

47.

\(h = 4t - 16t^2 ,~~\) for \(t\)

48.

\(P = IE - RI^2 ,~~\) for \(I\)

49.

\(s=vt-\dfrac{1}{2} at^2 ,~~\) for \(t\)

50.

\(S=\dfrac{n^2+n}{2} ,~~ \) for \(n\)

51.

\(3x^2 + xy + y^2 = 2 ,~~\) for \(y\)

52.

\(y^2 - 3xy + x^2 = 3 ,~~ \) for \(x\)

For Problems 53-60, solve for \(y\) in terms of \(x\text{.}\) Use whichever method of solution seems easiest.

53.

\(x^2 y - y^2 = 0\)

54.

\(x^2 y^2 - y = 0\)

55.

\((2y + 3x)^2 = 9\)

56.

\((3y - 2x)^2 = 4\)

57.

\(4x^2 - 9y^2 = 36\)

58.

\(9x^2 + 4y^2 = 36\)

59.

\(4x^2 - 25y^2 = 0\)

60.

\((2x - 5y)^2 = 0\)

For Problems 61-66, solve the formula for the indicated variable.

61.

\(V = \pi (r - 3)^2h , ~~ \) for \(r\)

62.

\(A = P(1 + r )^2 , ~~ \) for \(P\)

63.

\(E = \dfrac{1}{2} mv^2 + mgh , ~~ \) for \(v\)

64.

\(h = \dfrac{1}{2}gt^2 + dl , ~~ \) for \(t\)

65.

\(V = 2(s^2 + t^2)w , ~~ \) for \(t\)

66.

\(V = \pi(r ^2 + R^2)h , ~~ \) for \(R\)

67.

What is the sum of the two solutions of the quadratic equation \(ax^2 + bx + c = 0\text{?}\)

Hint

The two solutions are given by the quadratic formula.

68.

What is the product of the two solutions of the quadratic equation \(ax^2 + bx + c = 0\text{?}\)

Hint

Do not try to multiply the two solutions given by the quadratic formula! Think about the factored form of the equation

In Problems 69 and 70, we prove the quadratic formula.

69.

Complete the square to find the solutions of the equation \(x^2 + bx + c = 0\text{.}\) (Your answers will be expressions in \(b\) and \(c\text{.}\))

70.

Complete the square to find the solutions of the equation \(ax^2 + bx + c = 0\text{.}\) (Your answers will be expressions in \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\))