PCC SLC Math Resources

When the product of two numbers is zero, at least one of the numbers in the product must be zero. This is a property unique to zero; e.g., when two numbers multiply to, say, ten, not only is it not the case that one of the two numbers need be ten, the only restriction on the two numbers at all is that neither be zero.

When solving an equation of form:

the next step in the process is to state:

We then soleve the two new equations separately and state our solutions, or solution set. For example:

The solutions are \(-\frac{7}{4}\) and \(\frac{9}{2}\text{.}\) The solution set is \(\{-\frac{7}{4}, \frac{9}{2}\}\text{.}\)

We frequently have to perform some preliminary steps before invoking the zero-product property. Specifically:

1. Completely expand both sides of the equation.
2. Add and/or subtract to/from both sides of the equation so that one side of the equation is zero. The next step will be easier if make sure that the second degree term (\(ax^2\)) has a positive coefficient.
3. Factor the non-zero side of the equation.
4. We are now set to invoke the zero-product property.

Several examples follow.

Example: Use the zero-product property to solve \(4x^2=4x+15\text{.}\)

The solutions are \(\frac{5}{2}\) and \(-\frac{3}{2}\text{.}\) The solution set is \(\{\frac{5}{2}, -\frac{3}{2}\}\text{.}\)

Example: Use the zero-product property to solve \((x+6)(x-2)=-16\text{.}\)

The solution is \(-2\text{.}\) The solution set is \(\{-2\}\text{.}\)

Example: Use the zero-product property to solve \((x+4)(4x-5)=(7x+10)(3x-2)\text{.}\)

The solutions are \(0\) and \(-\frac{5}{17}\text{.}\) The solution set is \(\{0, -\frac{5}{17}\}\text{.}\)

Example: Use the zero-product property to solve \(2-x^2=(2-x)^2\text{.}\)

The solution is \(1\text{.}\) The solution set is \(\{1\}\text{.}\)

Click here to access some practice problems: Practice Exercises 1.12.2.1.

If \(k\) is a positive number and \(u^2=k\text{,}\) then:

The latter set of equations is frequently abbreviate as

which is read aloud as "\(u\) is equal to plus or minus the square root of \(k\text{.}\)"

For example, if \(x^2=64\text{,}\) then \(x= \pm \sqrt{64}\) which, of course, simplifies to \(x= \pm 8\text{.}\) So the original equation has two solutions: \(-8\) and \(8\text{.}\)

Another example follows. Solve \((3x+5)^2=16\text{.}\)

The solutions are \(-3\) and \(-\frac{1}{3}\text{.}\) The solution set is \(\{-3, -\frac{1}{3}\}\text{.}\)

We frequently have to do some preliminary work to have an equation of form \(u^2=k\text{.}\) The process involves completing the square. Completing the square entails adding a number to an expression of form \(x^2+bx\) so that the resultant trinomial factors into a perfect square. The number that complete the square is always the square of half of \(b\text{.}\)

For example, the number that completes the square for \(x^2+10x\) is \(5^2\) which is \(25\text{.}\) Note:

The steps followed when using the square root method are listed below.

1. Perform any and all manipulations so that the equation has the form \(x^2+bx=c\text{.}\)
2. Complete the square by adding \((\frac{b}{2})^2\text{.}\) Note that to keep the equation in balance \((\frac{b}{2})^2\) also needs to be added to the other side of the equation.
3. Factor the non-constant side of the equation. The equation should now have form \(u^2=k\text{.}\)
4. If the constant, \(k\text{,}\) is negative, then the equation has no solution. If \(k\) is zero, then the equation has one solution which is determined by solving \(u=0\text{.}\) Step 5 is written under the assumption that \(k\) is a positive number.
5. Invoke the square root property, i.e. \(u=\pm \sqrt{k}\text{.}\) If \(k\) is a perfect square, you need to write out and solve two equations.
   \begin{equation*} u=-\sqrt{k}\text{ or }u=\sqrt{k}\text{.} \end{equation*}
   If \(k\) is not a perfect square, you may solve for \(u\) maintaining the plus/minus sign. Make sure to simplify the square root should it simplify.
6. State your solutions and/or solution set.

Example: Solve \(x^2-6x-7=0\) by the square root method after first completing the square.

The solutions are \(-1\) and \(7\text{.}\) The solution set is \(\{-1, 7\}\text{.}\)

Example: Solve \((2x-1)(2x+1)=7-32x\) by the square root method after first completing the square.

The solutions are \(-4-3\sqrt{2}\) and \(-4+3\sqrt{2}\text{.}\)

The solutions set is \(\{-4-3\sqrt{2}, -4+3\sqrt{2}\}\text{.}\)

Example: Solve \(x^2+9x+25=0\) by the square root method after first completing the square.

The equation has no real number solutions.

Over the real numbers, the solution set is \(\emptyset\text{.}\)

Example: Solve \(4x^2-28x=-11\) by the square root method after first completing the square.

The solutions are \(\frac{7-\sqrt{38}}{2}\) and \(\frac{7+\sqrt{38}}{2}\text{.}\)

The solution set is \(\{\frac{7-\sqrt{38}}{2}, \frac{7+\sqrt{38}}{2}\}\text{.}\)

Click here to access some practice problems: Practice Exercises 1.12.2.2.

The solutions to a quadratic equation of form

can be found using the formula

The solution equation is called the quadratic formula.

For example, let's use the quadratic formula to solve the equation

We begin by identifying the constants \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)

We now replace the constants in the quadratic formula with their respective values and simplify the result.

The symbol "\(\pm\)" is read "plus or minus" and implies that there are two solutions - one when the symbol is executed as subtraction and another when the symbol is executed as addition. Picking up where we left off.

So the solutions to the equation are \(-4\) and \(8\) and the solution set is \(\{-4, 8\}\text{.}\)

As just demonstrated, one can use the quadratic formula without having an understanding of its origin. The remainder of this section is devoted to the derivation of the quadratic formula. If you'd rather just get on with using it, go ahead to the next section.

The derivation of the quadratic equation relies on the process of completing the square. We being with the equation

Before completing the square we need to isolate the constant and divide both sides by \(a\) so that the coefficient on the squared term is \(1\text{.}\)

We now complete the square on the left side of the equation by adding the square of half the linear coefficient. This action is balanced by adding the same expression to the right side of the equation. We'll then go ahead and combine the terms on the right side of the equation.

We now factor the left side into its perfect square and complete the derivation using the square root method. Note that when we simplify \(\sqrt{4a^2}\) we can simply state \(2a\text{.}\) It's true that we don't know whether \(a\) is a positive number or a negative number, but when we extract the roots we introduce the \(\pm\) sign, so we will have two roots, one positive and one negative, regardless of whether \(a\) is positive or negative (assuming that \(b^2-4ac \neq 0\)). If \(b^2-4ac=0\text{,}\) then \(\frac{b^2-4ac}{4a^2}=\frac{0}{2a}=0\) whether \(a\) is positive or negative.

Let's use the quadratic formula to solve the equation \(2x^2-5x=7\text{.}\)

We begin by making the right side of the equation zero and stating the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)

Finally, we state our solutions and solution set.

The solutions are \(-1\) and \(\frac{7}{2}\text{.}\) The solution set is \(\{-1,\frac{7}{2}\}\)

When the square root expression does not simplify to an integer, there's no need to spit the solution into two separate equations, although we do need to make sure that we state the solutions as separate numbers (assuming that there are two solutions). We also need to make sure that we completely simplify the square root and the fraction.

Let's use the quadratic formula to solve the equation \(x^2+10x+5=0\text{.}\)

The equation is already in standard form, so we can go ahead and state the values of \(a\text{,}\) \(b\text{,}\) and \(c\) and proceed with the quadratic formula.

The solutions are \(-5-2\sqrt{5}\) and \(-5+2\sqrt{5}\text{.}\)

The solution set is \(\{-5-2\sqrt{5}, -5+2\sqrt{5}\}\text{.}\)

When the ultimate value under the square root symbol is zero, the equation has only one solution. This is because \(\sqrt{0}=0\text{,}\) and it doesn't alter the result changing subtraction of zero to addition of zero.

When the ultimate value under the square root symbol is negative, the equation has no real number solutions because square roots of negative numbers are not real numbers (although they really are numbers). In many situations we are only interested in real number solutions, so we stop the solution process once the negative radicand has been identified and state that there are no real number solutions.

Difficult as it may be to believe, there are real life applications where numbers with imaginary parts (square roots of negative numbers) are meaningful. Because of this, we do need to examine some equations where the solutions are complex numbers with imaginary parts. We do this in the next section.

Click here to access some practice problems: Practice Exercises 1.12.2.3.

Let's solve the quadratic equation \(4x^2-12x+25=0\text{.}\) The equation is already in standard form, so we can state the values of \(a\text{,}\) \(b\text{,}\) and \(c\) and proceed with the quadratic formula.

If we were only concerned with real number solutions we could state our conclusion now, because \(\sqrt{-256}\) is not a real number. However, let's assume that we our also interested in solutions that include imaginary parts and proceed. In our stated solutions we want the real part of each number separate from the imaginary part, so we will simplify towards that end.

The solutions are \(\frac{3}{2}-2i\) and \(\frac{3}{2}+2i\text{.}\)

The solution set is \(\{\frac{3}{2}-2i,\frac{3}{2}+2i\}\text{.}\)

Let's see another example.

Use the quadratic formula to solve \(x(2x-5)=4(x^2+3)\text{.}\)

We need to manipulate the equation into standard form before we state the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) I prefer that \(a\) is a positive number, so after I expand both sides I am going to move all the terms to the right side of the equation.

The solutions are \(-\frac{5}{4}-\frac{\sqrt{71}}{4}i\) and \(-\frac{5}{4}+\frac{\sqrt{71}}{4}i\text{.}\)

The solution set is \(\{-\frac{5}{4}-\frac{\sqrt{71}}{4}i,-\frac{5}{4}+\frac{\sqrt{71}}{4}i\}\text{.}\)

Click here to access some practice problems: Practice Exercises 1.12.2.4.

There are times when we are concerned about the nature of the solutions to a quadratic equation rather than the precise solutions themselves. For example, when the equation \(ax^2+bx+c=0\) has two, one, or zero real number solutions, we know that the graph of \(y=ax^2+bx+c\) has, respectively, two, one, or zero \(x\)-intercepts.

Consider the quadratic formula.

Because \(\sqrt{b^2-4ac}\) is the expression that is both added and subtracted, and it the part of the expression that can be an imaginary number, it is our focus when determining the nature of the solution set. We can narrow our focus even further, because the nature is determined by the value of \(b^2-4ac\text{.}\) When that value is positive, the equation will have two real number solutions. When the value is zero there will be one real number solution, and when the value is negative there will be two complex number solutions each with non-zero imaginary coefficients.

Because the expression \(b^2-4ac\) plays the key role in the nature of the solution set to a quadratic equation, it is given a name. The expression is called the determinant. Let's consider three examples.

Example

For the equation \(x^2-4x-7=0\text{,}\) \(a=1\text{,}\) \(b=-4\text{,}\) and \(c=-7\text{.}\) The discriminant is derived below.

Because \(b^2-4a \gt 0\text{,}\) we know that the equation \(x^2-4x-7=0\) has two real number solutions and that the graph of \(y=x^2-4-7\) has two \(x\)-intercepts.

Example

For the equation \(2x^2-6x+9=0\text{,}\) \(a=2\text{,}\) \(b=-6\text{,}\) and \(c=9\text{.}\) The discriminant is derived below.

Because \(b^2-4ac \lt 0\text{,}\) we know that the equation \(2x^2-6x+9=0\) has no real number solutions and that the graph of \(y=2x^2-6x+9\) has no \(x\)-intercepts. The equation \(2x^2-6x+9=0\) has two complex number solutions, each of which has a non-zero imaginary coefficient.

Example

For the equation \(4x^2-12x+9=0\text{,}\) \(a=4\text{,}\) \(b=-12\text{,}\) and \(c=9\text{.}\) The discriminant is derived below.

Because \(b^2-4ac=0\text{,}\) we know that the equation \(4x^2-12x+9=0\) has exactly one real number solution and that the graph of \(y=4x^2-12x+9\) has exactly one \(x\)-intercept.

Click here to access some practice problems: Practice Exercises 1.12.2.5.

Example

You may be familiar with the Pythagorean Theorem: If the legs of a right triangle have lengths of \(a\) and \(b\) while the hypotenuse has a length of \(c\text{,}\) then

Suppose that we have a right triangle where one leg is \(1\) inch longer than the other leg and the hypotenuse is \(1\) inch longer than the longer leg. Let's determine the length of each side of the triangle.

Let \(x\) represent the length (in) of the shorter leg of the triangle. Then the length (in) of the longer leg is represented by \(x+1\) and the length (in) of the hypotenuse is represented by \(x+2\text{.}\)

Applying the Pythagorean Theorem we have

We begin solving the equation by expanding all expressions in the equation.

Since a length cannot be negative, we reject the negative solution. So the lengths of the triangle are \(3\) inches, \(4\) inches, and \(5\) inches.

Example

The area formula for a trapezoid is

where \(b_1\) and \(b_2\) are the lengths of the parallel sides of the trapezoid (called bases) and \(h\) is the perpendicular length between the two bases (called the height).

Suppose that we have a trapezoid where the length of one of the bases is twice the length of the other base and the height is five inches shorter than the longer base. Suppose further that the area of the trapezoid is \(132\) \(\text{in}^2\text{.}\) Let's determine the length of each of the bases and the height.

Let's define \(x\) to be the length (in) of the shorter base. Then the length (in) of the longer base is represented by \(2x\) and the length (in) of the height is represented by \(2x-5\text{.}\)

Using the area formula for a trapezoid we have:

We begin solving the equation by clearing the fraction and expanding the left side of the equation.

Since length cannot be negative, we reject the negative solution. So the lengths of the two bases are \(8\) inches and \(16\) inches while the length of the height is \(11\) inches.

Example

If an object is dropped from a height (feet) of \(h\text{,}\) then the distance between the object and the ground after \(t\) seconds can be approximated by the function \(f(t)=-16t^2+h\text{.}\) Let's use this information to approximate the amount of time it takes an object to reach the ground after being dropped from a height of \(500\) feet.

Our distance function is \(f(t)=-16t^2+500\text{.}\) When the object reaches the ground the value of \(f\) is \(0\text{.}\) So the equation we need to solve is

We can quickly determine the solution by isolating \(t^2\) and employing the square root method.

Since the object cannot reach the ground before it has been released, we reject the negative solution. Using a calculator we have:

So it takes about 5.59 seconds for the object to reach the ground after its release.

Click here to access some practice problems: Practice Exercises 1.12.2.6.

Use the zero-product property to solve each quadratic equation. State the solutions to each equation as well as the solution set to each equation.

1. \((3x+8)(5x-7)=0\)
2. \(x(x-6)=0\)
3. \(x(x-3)=10\)
4. \(x^2+12x=-36\)
5. \(4t(8t+9)=5\)
6. \(2y^2=y\)
7. \((w-10)(w+1)=-10\)

Use the square root method to solve each of the following quadratic equations over the real numbers. Make sure that all solutions are completely simplified. State the solutions to each equation as well as the solution set to each equation.

1. \((2t-1)^2=9\)
2. \((3x-2)^2=72\)
3. \(x^2-8x+16=12\)
4. \(y^2+20y+80=0\)
5. \(x^2+4x+7=0\)
6. \(w^2-7w+7=0\)

Use the quadratic formula to solve each of the following equations over the real numbers. For each equation state both the solutions and the solution set.

1. \(x^2+6x-4=0\)
2. \(t^2+10t=39\)
3. \(3x^2-2x+5=0\)
4. \(2x(3x-5)=5\)
5. \(1-4x^2=x^2+10x+6\)

Use the square root method to solve each of the following equations over the complex numbers. Make sure that all solutions have been completely simplified. State both the solutions and the solution set.

1. \((x+6)^2=-9\)
2. \((3x-8)^2=-20\)
3. \(x^2-12x+52=0\)
4. \(3w^2-12w+24=0\)

Use the quadratic formula to solve each of the following equations over the complex numbers. Make sure that all solutions have been completely simplified. State both the solutions and the solution set.

1. \(4x^2+2x+1=0\)
2. \(t^2-3t+5=0\)
3. \(3x^2+16=0\)
4. \(y^2-4y+9=0\)

For each equation, use the value of the discriminant to determine the nature of the solution. That's all - do not find the actual solutions.

1. \(3x^2+4x-7=0\)
2. \(36-12x-x^2=0\)
3. \(x^2+4=0\)
4. \(3x(3x-16)=-64\)
5. \(7x^2=15x\)

1. A bean bag is shot into the air from the top of a 168 ft building with an initial upward velocity of \(88\) ft/s. It can shown (using calculus) that the height (ft) of the bag relative to the ground is given by the function \(h(t)=-16t^2+88t+168\) where \(t\) is the number of seconds that have elapsed since the bean bag was released. Determine the amount of time it takes for the bean bag to reach the ground upon its release.
   \begin{equation*} \end{equation*}
2. One square inside another, with a \(2\) meter gap between the walls of the smaller square and larger square. The area of the larger square is \(484 \text{m}^2\text{.}\) Determine the perimeter of the smaller square.
   \begin{equation*} \end{equation*}
3. One leg of a right triangle is \(7\) cm longer than the other and the hypotenuse of the triangle is \(1\) cm longer than the longer leg of the triangle. Determine the length of each side of the triangle.
   \begin{equation*} \end{equation*}

Follow this link to see some examples: Written Examples 1.12.1.1.

Use the zero-product property to solve each quadratic equation.

1. \((2x+5)(3x-4)=0\)
2. \(4x(x+6)=0\)
3. \(x^2-10x-24=0\)
4. \(x(x+7)=30\)
5. \(12w^2=-9w\)
6. \(25y^2+30y+9=0\)

Follow this link to see some examples: Written Examples 1.12.1.2.

Use the square root method to solve each of the following quadratic equations. Make sure that all solutions are completely simplified. Note that for some of the equations you will need to first complete the square.

1. \((5t+7)^2=49\)
2. \((2x-3)^2=12\)
3. \(x^2+8x-20=0\)
4. \(y^2-20y+40=0\)

Follow this link to see some examples: Written Examples 1.12.1.4.

Use the quadratic formula to solve each of the following equations over the real numbers.

1. \(2x^2+8x-5=0\)
2. \(t^2-15t=100\)
3. \(-x^2+4x+3=0\)
4. \(x(3x-5)=-6\)

Follow this link to see some examples: Written Examples 1.12.1.5.

Use the square root method to solve each of the following equations. Make sure that all solutions have been completely simplified. Note that for one of the equations you will need to first complete the square.

1. \((x-7)^2=-1\)
2. \((2x-9)^2=-48\)
3. \(-2w^2+12w-24=0\)

Use the quadratic formula to solve each of the following equations. Make sure that all solutions have been completely simplified. State both the solutions and the solution set.

1. \(6x^2+1=2x\)
2. \(y^2-18y+97=0\)

Follow this link to see some examples: Written Examples 1.12.1.6.

For each equation, use the value of the discriminant to determine the nature of the solution. That's all - do not find the actual solutions.

1. \(12x^2+4x+3=0\)
2. \(-x^2+4x+4=0\)
3. \(-49+14x-x^2=0\)