Quadratic Equations
Learning Objectives
- Recognize a quadratic equation
- Use the zero product principle to solve quadratic equations that can be factored
- Identify solutions to quadratic equations on a graph
- Square Roots and Completing the Square
- Use the square root property to solve a quadratic equation
- Complete the square to solve a quadratic equation
- The Quadratic Formula
- Write a quadratic equation in standard form and identify the values of a, b, and c in a standard form quadratic equation.
- Use the Quadratic Formula to find all real solutions of a quadratic equation, recognize when there are no real solutions
- Solve application problems involving quadratic equations
The Zero-Product Property and Quadratic Equations
The zero-product property statesSolving 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.Reminder: 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
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].
Example
Solve the difference of squares equation using the zero-product property: [latex]{x}^{2}-9=0[/latex].Answer: Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.
Solving Quadratics with a Leading Coefficient of [latex]\ne1[/latex]
Recall that when the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:- With the quadratic in standard form, [latex]a{x}^{2}+bx+c=0[/latex], multiply [latex]a\cdot c[/latex].
- Find two numbers whose product equals [latex]ac[/latex] and whose sum equals [latex]b[/latex].
- Rewrite the equation replacing the [latex]bx[/latex] term with two terms using the numbers found in step 1 as coefficients of x.
- Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
- Factor out the expression in parentheses.
- Set the expressions equal to zero and solve for the variable.
Example
Use grouping to factor and solve the quadratic equation: [latex]4{x}^{2}+15x+9=0[/latex].Answer: First, multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].
Example
Solve the equation by factoring: [latex]-3{x}^{3}-5{x}^{2}-2x=0[/latex].Answer: This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out [latex]-x[/latex] from all of the terms and then proceed with grouping.
Solve a Quadratic Equation by the Square Root Property
One way to solve the quadratic equation [latex]x^{2}=9[/latex] is to subtract 9 from both sides to get one side equal to 0: [latex]x^{2}-9=0[/latex]. The expression on the left can be factored, it is a difference of squares: [latex]\left(x+3\right)\left(x–3\right)=0[/latex]. Using the zero factor property, you know this means [latex]x+3=0[/latex] or [latex]x–3=0[/latex], so [latex]x=−3[/latex] or 3. Another property would let you solve that equation more easily is called the square root property.The Square Root Property
If [latex]x^{2}=a[/latex], then [latex] x=\sqrt{a}[/latex] or [latex] -\sqrt{a}[/latex]. The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of a and the negative square root of a.Example
Solve using the Square Root Property. [latex]x^{2}=9[/latex]Answer: Since one side is simply [latex]x^{2}[/latex], you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!
[latex]\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}[/latex]
Answer
[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])Example
Solve. [latex]10x^{2}+5=85[/latex]Answer: If you try taking the square root of both sides of the original equation, you will have [latex] \sqrt{10{{x}^{2}}+5}[/latex] on the left, and you can’t simplify that. Subtract 5 from both sides to get the [latex]x^{2}[/latex] term by itself.
[latex]10x^{2}+5=85[/latex]
You could now take the square root of both sides, but you would have [latex] \sqrt{10}[/latex] as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.[latex]10x^{2}=80[/latex]
Now you have only [latex]x^{2}[/latex] on the left, so you can use the Square Root Property easily. Be sure to simplify the radical if possible.[latex] \begin{array}{l}{{x}^{2}}=8\\\,\,\,x=\pm \sqrt{8}\\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]
Answer
[latex-display] x=\pm 2\sqrt{2}[/latex-display]Example
Solve. [latex]\left(x–2\right)^{2}–50=0[/latex]Answer: Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.
[latex]\left(x-2\right)^{2}-50=0[/latex]
Because [latex]\left(x–2\right)^{2}[/latex] is a squared quantity, you can take the square root of both sides.[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}[/latex]
To isolate x on the left, you need to add 2 to both sides. Be sure to simplify the radical if possible.[latex] \begin{array}{l}x=2\pm \sqrt{50}\\\,\,\,\,=2\pm \sqrt{(25)(2)}\\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]
Answer
[latex-display] x=2\pm 5\sqrt{2}[/latex-display] In the next video you will see more examples of using square roots to solve quadratic equations. https://youtu.be/4H5qZ_-8YM4Solve a Quadratic Equation by Completing the Square
Not all quadratic equations can be factored or solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. First, let's make sure we can recognize a perfect square trinomial and how factor it.Example
Factor [latex]9x^{2}–24x+16[/latex].Answer: First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares. [latex-display]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array}[/latex-display] Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms. [latex-display]24x=2\left(3x\right)\left(4\right)[/latex-display] A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as [latex](r–s)^{2}[/latex]. In this case, the middle term is subtracted, so subtract r and s and square it to get [latex](r–s)^{2}[/latex]. [latex-display]\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex-display]
Steps for Completing The Square
We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.- Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right side of the equal sign.
[latex]{x}^{2}+4x=-1[/latex]
- Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
[latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
- Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
[latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
- The left side of the equation can now be factored as a perfect square.
[latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
- Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
- The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].
Example
Solve by completing the square. [latex]x^{2}–12x–4=0[/latex]Answer: Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation. First, move the constant term to the right side of the equal sign.Identify b.
[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]
Then, take [latex]\frac{1}{2}[/latex] of the b term and square it. Add [latex] {{\left( \frac{b}{2}\right)}^{2}}[/latex] to complete the square, so [latex] {{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36[/latex]. Add the value to both sides of the equation and simplify.[latex]\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}[/latex]
Rewrite the left side as a squared binomial.[latex]\left(x-6\right)^{2}=40[/latex]
Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.[latex] x-6=\pm\sqrt{40}[/latex]
Solve for x by adding 6 to both sides. Simplify as needed.[latex] \begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]
Answer
[latex-display] x=6\pm 2\sqrt{10}[/latex-display]Example
Solve by completing the square: [latex]{x}^{2}-3x - 5=0[/latex].Answer: First, move the constant term to the right side of the equal sign.
Example
Solve by completing the square. [latex]x^{2}+16x+17=-47[/latex].Answer: Rewrite the equation so the left side has the form [latex]x^{2}+bx[/latex]. Identify b.
[latex]\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}[/latex]
Add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex], which is [latex] {{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64[/latex], to both sides.[latex]\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}[/latex]
Write the left side as a squared binomial.[latex]\left(x+8\right)^{2}=0[/latex]
Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.[latex]x+8=0[/latex]
Solve for x.[latex]x=-8[/latex]
Answer
[latex-display]x=-8[/latex-display]The Quadratic Formula
You can solve any quadratic equation by completing the square—rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[/latex] and then solve for x, you find that [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. This equation is known as the Quadratic Formula. We can derive the quadratic formula by completing the square. First, assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[/latex] and obtain a positive a. Given [latex]a{x}^{2}+bx+c=0[/latex], [latex]a\ne 0[/latex], we will complete the square as follows:- First, move the constant term to the right side of the equal sign:
[latex]a{x}^{2}+bx=-c[/latex]
- As we want the leading coefficient to equal 1, divide through by a:
[latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
- Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign:
[latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
- Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
[latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
- Now, use the square root property, which gives
[latex]\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
- Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus,
[latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]
Solving a Quadratic Equation using the Quadratic Formula
The Quadratic Formula will work with any quadratic equation, but only if the equation is in standard form, [latex]ax^{2}+bx+c=0[/latex]. To use it, follow these steps.- Put the equation in standard form first.
- Identify the coefficients, a, b, and c. Be careful to include negative signs if the bx or c terms are subtracted.
- Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
- Simplify as much as possible.
- Use the [latex]\pm[/latex] in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted.
- Simplify both values to get the possible solutions.
Example
Use the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[/latex].Answer: First write the equation in standard form.
[latex]\begin{array}{r}x^{2}+4x=5\,\,\,\\x^{2}+4x-5=0\,\,\,\\\\a=1,b=4,c=-5\end{array}[/latex]
Note that the subtraction sign means the constant c is negative.[latex] \begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,4x\,\,\,-\,\,\,5\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex]
Substitute the values into the Quadratic Formula. [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\[/latex][latex] \begin{array}{l}\\x=\frac{-4\pm \sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\end{array}[/latex]
Simplify, being careful to get the signs correct.[latex]x=\frac{-4\pm\sqrt{16+20}}{2}[/latex]
Simplify some more.[latex] x=\frac{-4\pm \sqrt{36}}{2}[/latex]
Simplify the radical: [latex] \sqrt{36}=6[/latex].[latex] x=\frac{-4\pm 6}{2}[/latex]
Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.[latex]\begin{array}{c}x=\frac{-4+6}{2}=\frac{2}{2}=1\\\\\text{or}\\\\x=\frac{-4-6}{2}=\frac{-10}{2}=-5\end{array}[/latex]
Answer
[latex-display]x=1\,\,\,\text{or}\,\,\,-5[/latex-display][latex]\begin{array}{r}x=1\\x^{2}+4x=5\\\left(1\right)^{2}+4\left(1\right)=5\\1+4=5\\5=5\end{array}[/latex] | [latex]\begin{array}{r}x=-5\\x^{2}+4x=5\,\,\,\,\,\\\left(-5\right)^{2}+4\left(-5\right)=5\,\,\,\,\,\\25-20=5\,\,\,\,\,\\5=5\,\,\,\,\,\end{array}[/latex] |
Example
Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[/latex].Answer: Subtract 6x from each side and add 16 to both sides to put the equation in standard form.
[latex]\begin{array}{l}x^{2}-2x=6x-16\\x^{2}-2x-6x+16=0\\x^{2}-8x+16=0\end{array}[/latex]
Identify the coefficients a, b, and c. [latex]x^{2}=1x^{2}[/latex], so [latex]a=1[/latex]. Since [latex]8x[/latex] is subtracted, b is negative. [latex]a=1,b=-8,c=16[/latex][latex] \begin{array}{r}{{x}^{2}}\,\,\,-\,\,\,8x\,\,\,+\,\,\,16\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,\,c\,\,\,\,=\,\,\,0\end{array}[/latex]
Substitute the values into the Quadratic Formula.[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\end{array}[/latex]
Simplify.[latex] x=\frac{8\pm \sqrt{64-64}}{2}[/latex]
Since the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.[latex] x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]
Answer
[latex-display]x=4[/latex-display][latex]\begin{array}{r}x^{2}-2x=6x-16\,\,\,\,\,\\\left(4\right)^{2}-2\left(4\right)=6\left(4\right)-16\\16-8=24-16\,\,\,\,\,\,\\8=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
In the next example, we will show that some quadratic equations do not have real solutions. As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.Example
Use the Quadratic Formula to solve the equation [latex]x^2+x=-x-3[/latex]Answer: Add x to both sides, and add 3 to both sides to get the quadratic equation in standard form.
[latex]\begin{array}{l}x^{2}+x=-x-3\\x^{2}+2x+3=0\end{array}[/latex]
Identify a, b, c.[latex]a=1, b=2, c=3[/latex]
Substitute values for a, b, c into the quadratic formula.
[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-2\pm \sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\end{array}[/latex]
Simplify
[latex] x=\frac{-2\pm \sqrt{-8}}{2}[/latex]
Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.
Answer
There are no real solutions.Applying the Quadratic Formula
Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge. A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation isn't completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.Example
A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called initial velocity) is [latex]−10[/latex] feet per second. (The negative value means it's heading toward the ground.) The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?Answer: When the ball hits the ground, the height is 0. Substitute 0 for h.
[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]
This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].[latex] t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]
Simplify. Be very careful with the signs.[latex] \begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]
Use a calculator to find both roots.t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].
Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.Answer
The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown. In the next video we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground. https://youtu.be/RcVeuJhcuL0Example
Bob made a quilt that is 4 ft [latex]\times[/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)Answer: Sketch the problem. Since you don’t know the width of the border, you will let the variable x represent the width. In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines. Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex]. (Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.) You are only interested in the area of the border strips. Write an expression for the area of the border.
Area of border = Area of the blue rectangle minus the area of the red rectangle
Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]
There are 10 sq ft of fabric for the border, so set the area of border to be 10.[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]
Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].[latex]10=20+8x+10x+4x^{2}–20[/latex]
Simplify.[latex]10=18x+4x^{2}[/latex]
Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}-10\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]
Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].[latex]\begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]
Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]
Simplify.[latex] x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]
Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.[latex]\begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]
Ignore the solution [latex]x=−5[/latex], since the width could not be negative.Answer
The width of the border should be 0.5 ft. Our last video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt. https://youtu.be/Zxe-SdwutxASummary
Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations, or any other quadratic equation! The Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex], is found by completing the square of the quadratic equation [latex] [/latex]. When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.Summary
You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or both a and b are 0. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation. Not all solutions are appropriate for some applications. In many real-world situations, negative solutions are not appropriate and must be discarded. Completing the square is used to change a binomial of the form [latex]x^{2}+bx[/latex] into a perfect square trinomial [latex] {{x}^{2}}+bx+{{\left( \frac{b}{2} \right)}^{2}}[/latex], which can be factored to [latex] {{\left( x+\frac{b}{2} \right)}^{2}}[/latex]. When solving quadratic equations by completing the square, be careful to add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex] to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for x. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.Licenses & Attributions
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- Quadratic Formula Application - Time for an Object to Hit the Ground. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Quadratic Formula Application - Determine the Width of a Border. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
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- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at : http://cnx.org/contents/[email protected]:1/Preface.
- Ex: Solve a Quadratic Equation Using Factor By Grouping. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex: Factor and Solve Quadratic Equation - Trinomial a = -1. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
- Ex 1: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex 2: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: Public Domain: No Known Copyright.
- Ex 1: Completing the Square - Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) . License: Public Domain: No Known Copyright.
- Ex 2: Completing the Square - Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex2: Quadratic Formula - Two Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex: Quadratic Formula - Two Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.