Learning Outcomes
- Write a quadratic equation in standard form and identify the values of[latex]a[/latex],[latex]b[/latex], and [latex]c[/latex] in astandard form quadratic equation.
- Use the Quadratic Formula to find all real solutions of a quadratic equation
Quadratic formula
- 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}\cdot\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]
Writing a Quadratic Equation in Standard Form
Remember that the form [latex]ax^{2}+bx+c=0[/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it isvital that you be sure the equation is in this form. If you do not, you might use the wrong values for a, b, or c, and then the formula will give incorrect solutions.The following examples show how to ensure that your quadratic equation is in standard form and then correctly identify the values you will be using for a, b, and c in the Quadratic Formula.Example
Rewrite the equation [latex]3x+2x^{2}+4=5[/latex]in standard form and identify a, b, and c.
Answer: First be sure that the right side of the equation is 0. In this case, all you need to do is subtract [latex]5[/latex] from both sides.
[latex]\begin{array}{c}3x+2x^{2}+4=5\\3x+2x^{2}+4–5=5–5\end{array}[/latex]
Simplify, and write the terms with the exponent on the variable in descending order.
[latex]\begin{array}{r}3x+2x^{2}-1=0\\2x^{2}+3x-1=0\end{array}[/latex]
Now that the equation is in standard form, you can read the values of a, b, and c from the coefficients and constant. Note that since the constant 1 is subtracted, c must be negative.
[latex]\begin{array}{l}2x^{2}\,\,\,+\,\,\,3x\,\,\,-\,\,\,1\,\,\,=\,\,\,0\\\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\\,ax^{2}\,\,\,\,\,\,\,\,\,\,bx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\\\\,\,a=2,\,\,b=3,\,\,c=-1\end{array}[/latex]
Answer
[latex-display]2x^{2}+3x–1=0;a=2,b=3,c=−1[/latex-display]
Example
Rewrite the equation [latex]2(x+3)^{2}–5x=6[/latex]in standard form and identify a, b, and c.
Answer: First be sure that the right side of the equation is [latex]0[/latex].
[latex]\begin{array}{c}2\left(x+3\right)^{2}–5x=6\\2(x+3)^{2}–5x–6=6–6\end{array}[/latex]
Expand the squared binomial, then simplify by combining like terms.Be sure to write the terms with the exponent on the variable in descending order.
[latex]\begin{array}{r}2\left(x^{2}+6x+9\right)-5x-6=0\\2x^{2}+12x+18–5x–6=0\\2x^{2}+12x–5x+18–6=0\\2x^{2}+7x+12=0\end{array}[/latex]
Now that the equation is in standard form, you can read the values of a, b, and c from the coefficients and constant.
[latex]\begin{array}{l}2x^{2}\,\,\,+\,\,\,7x\,\,\,+\,\,\,12\,\,\,=\,\,\,0\\\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\\\\,\,\,\,\,\,a=2,\,\,b=7,\,\,c=7\end{array}[/latex]
Answer
[latex-display]2x^{2}+7x+12=0;\,\,a=2,b=7,c=12[/latex-display]
Try It
[ohm_question]91600[/ohm_question]
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 sure to include negative signs if the bx or c terms are subtracted.
- Carefully substitute the values noted in step[latex]2[/latex] 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,[latex]6[/latex] 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]
The solutions are [latex]x=1\,\,\,\text{or}\,\,\,-5[/latex].
| [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] |
Try it
[ohm_question]4015[/ohm_question]
Example
Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[/latex].
Answer:Subtract [latex]6[/latex]x from each side and add[latex]16[/latex] 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[latex]0[/latex] is[latex]0[/latex], and both adding and subtracting[latex]0[/latex] give the same result, there is only one possible value.
[latex] x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]
The answer is [latex]x=4[/latex].
[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 following video we show an example of using the quadratic formula to solve a quadraticequation that has one repeated solution.https://youtu.be/OXwwzWcxFgEIn 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 [latex]x[/latex] 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.
Summary
The Quadratic Formula is a useful way to solve any 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.Contribute!
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Licenses & Attributions
CC licensed content, Original
- 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.
CC licensed content, Shared previously
- 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) for Lumen Learning. License: CC BY: Attribution.
