Factoring and the Square Root Property
Learning Objectives
- Factor a quadratic equation to solve it
- Use the square root property to solve a quadratic equation
- Use the pythagorean theorem and the square root property to find the unknown length of a triangle leg
[latex]\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&={x}^{2}+3x - 2x - 6\hfill \\ \hfill&={x}^{2}+x - 6\hfill \end{array}[/latex]
The product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[/latex], where a, b, and c are real numbers, and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] is in standard form.
We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.
A General Note: The Zero-Product Property and Quadratic Equations
The zero-product property states[latex]\text{If }a\cdot b=0,\text{ then }a=0\text{ or }b=0[/latex],
where a and b are real numbers or algebraic expressions.
A quadratic equation is an equation containing a second-degree polynomial; for example
[latex]a{x}^{2}+bx+c=0[/latex]
where a, b, and c are real numbers, and if [latex]a\ne 0[/latex], it is in standard form.
Solving Quadratics with a Leading Coefficient of 1
In the quadratic equation [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], is 1. We have one method of factoring quadratic equations in this form.How To: Given a quadratic equation with the leading coefficient of 1, factor it.
- Find two numbers whose product equals c and whose sum equals b.
- Use those numbers to write two factors of the form [latex]\left(x+k\right)\text{ or }\left(x-k\right)[/latex], where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and [latex]-2[/latex], the factors are [latex]\left(x+1\right)\left(x - 2\right)[/latex].
- Solve using the zero-product property by setting each factor equal to zero and solving for the variable.
Example: Factoring and Solving a Quadratic with Leading Coefficient of 1
Factor and solve the equation: [latex]{x}^{2}+x - 6=0[/latex].Answer: To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[/latex].
[latex]\begin{array}{l}1\cdot \left(-6\right)\hfill \\ \left(-6\right)\cdot 1\hfill \\ 2\cdot \left(-3\right)\hfill \\ 3\cdot \left(-2\right)\hfill \end{array}[/latex]
The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
[latex]\left(x - 2\right)\left(x+3\right)=0[/latex]
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
[latex]\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&=0\hfill \\ \left(x - 2\right)\hfill&=0\hfill \\ x\hfill&=2\hfill \\ \left(x+3\right)\hfill&=0\hfill \\ x\hfill&=-3\hfill \end{array}[/latex]
The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex]. We can see how the solutions relate to the graph below. The solutions are the x-intercepts of [latex]{x}^{2}+x - 6=0[/latex].
Try It
Factor and solve the quadratic equation: [latex]{x}^{2}-5x - 6=0[/latex].Answer: [latex]\left(x - 6\right)\left(x+1\right)=0;x=6,x=-1[/latex]
Using the Pythagorean Theorem
One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], where [latex]a[/latex] and [latex]b[/latex] refer to the legs of a right triangle adjacent to the [latex]90^\circ [/latex] angle, and [latex]c[/latex] refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side. The Pythagorean Theorem is given as[latex]{a}^{2}+{b}^{2}={c}^{2}[/latex]
where [latex]a[/latex] and [latex]b[/latex] refer to the legs of a right triangle adjacent to the [latex]{90}^{\circ }[/latex] angle, and [latex]c[/latex] refers to the hypotenuse.
Figure 4
Example: Finding the Length of the Missing Side of a Right Triangle
Find the length of the missing side of the right triangle.Answer: As we have measurements for side b and the hypotenuse, the missing side is a.
[latex]\begin{array}{l}{a}^{2}+{b}^{2}\hfill&={c}^{2}\hfill \\ {a}^{2}+{\left(4\right)}^{2}\hfill&={\left(12\right)}^{2}\hfill \\ {a}^{2}+16\hfill&=144\hfill \\ {a}^{2}\hfill&=128\hfill \\ a\hfill&=\sqrt{128}\hfill \\ \hfill&=8\sqrt{2}\hfill \end{array}[/latex]
Try It
Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.Answer: [latex]5[/latex] units
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