10.6: Solving Quadratic Equations Using the Quadratic Formula
- Page ID
- 49406
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Standard Form Of A Quadratic Equation
We have observed that a quadratic equation is an equation of the form
\(ax^2 + bx + c = 0, a \not = 0\).
where
- \(a\) is the coefficient of the quadratic term,
- \(b\) is the coefficient of the linear term,
- \(c\) is the constant term.
The equation \(ax^2 + bc + c = 0\) is the standard form of a quadratic equation.
Sample Set A
Determine the values of \(a, b\), and \(c\).
In the equation \(3x^2 + 5x + 2 = 0\),
\(a = 3\)
\(b = 5\)
\(c = 2\)
In the equation \(12x^2-2x-1 = 0\),
\(a = 12\)
\(b = -2\)
\(c = -1\)
In the equation \(2y^2 + 3 = 0\),
\(a = 2\)
\(b = 0\) -> Because the equation could be wrirten \(2y^2 + 0y + 3 = 0\)
\(c = 3\)
In the equation \(-8y^2 + 11y = 0\),
\(a = -8\)
\(b = 11\)
\(c = 0\) -> Since \(-8y^2 + 11y + 0 = 0\)
In the equation \(z^2 = z + 8\),
\(a = 1\)
\(b = -1\)
\(c = -8\)
When we write the equation in standard form, we get \(z^2 - z - 8 = 0\)
Practice Set A
Determine the values of \(a, b\), and \(c\) in the following quadratic equations.
\(4x^2 - 3x + 5 = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= 4\\
b &= -3\\
c &= 5
\end{array}\)
\(3y^2 - 2y + 9 = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= 3\\
b &= -2\\
c &= 9
\end{array}\)
\(x^2 - 5x - 1 = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= 1\\
b &= -5\\
c &= -1
\end{array}\)
\(x^2 - 4 = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= 1\\
b &= 0\\
c &= -4
\end{array}\)
\(x^2 - 2x = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= 1\\
b &= -2\\
c &= 0
\end{array}\)
\(y^2 = 5y - 6\)
- Answer
-
\(\begin{array}{flushleft}
a &= 1\\
b &= -5\\
c &= 6
\end{array}\)
\(2x^2 - 4x = -1\)
- Answer
-
\(\begin{array}{flushleft}
a &= 2\\
b &= -4\\
c &= 1
\end{array}\)
\(5x - 3 = -3x^2\)
- Answer
-
\(\begin{array}{flushleft}
a &= 3\\
b &= 5\\
c &= -3
\end{array}\)
\(2x - 11 - 3x^2 = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= -3\\
b &= 2\\
c &= -11
\end{array}\)
\(y^2 = 0\)
- Answer
-
\(\begin{array}{flushleft}
a &= 1\\
b &= 0\\
c &= 0
\end{array}\)
The solutions to all quadratic equations depend only and completely on the values \(a, b\) and \(c\)
The Quadratic Formula
When a quadratic equation is written in standard form so that the values \(a, b\), and \(c\) are readily determined, the equation can be solved using the quadratic formula. The values that satisfy the equation are found by substituting the values \(a, b\), and \(c\) into the formula
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Keep in mind that the plus or minus symbol, \pm, is just a shorthand way or denoting the two possibilities:
\(x = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}\) and \(x = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}\)
The quadratic formula can be derived by using the method of completing the square.
Derivation Of The Quadratic Formula
Solve \(ax^2 + bx = -c = 0\) for \(x\) by completing the square.
Subtract \(c\) from both sides.
\(ax^2 + bx = -c\).
Divide both sides by \(a\), the coefficient of \(x^2\).
\(x^2 + \dfrac{b}{a}x = \dfrac{-c}{a}\)
Now we have the proper form to complete the square. Take one half the coefficient of \(x\), square it, and add the result to both sides of the equation found in step 2.
a) \(\dfrac{1}{2} \cdot \dfrac{b}{a} = \dfrac{b}{2a}\) is one half the coefficient of \(x\).
b) \((\dfrac{b}{2a})^2\) is the square of one half the coefficient of \(x\)
\(x^2 + \dfrac{b}{a} + (\dfrac{b}{2a})^2 = \dfrac{-c}{a} + (\dfrac{b}{2a})^2\)
The left side of the equation is now a perfect square trinomial and can be factored. This gives us:
\((x + \dfrac{b}{2a})^2 = \dfrac{-c}{a} + \dfrac{b^2}{4a^2}\)
Add the two fractions on the right side of the equation. The LCD \(= 4a^2\).
\(\begin{array}{flushleft}
(x + \dfrac{b}{2a})^2 &= \dfrac{-4ac}{4a^2} + \dfrac{b^2}{4a^2}\\
(x + \dfrac{b}{2a})^2 &= \dfrac{-4ac + b^2}{4a^2}\\
(x + \dfrac{b}{2a})^2 &= \dfrac{b^2 - 4ac}{4a^2}
\end{array}\)
Solve for \(x\) using the method of extraction of roots.
\(\begin{array}{flushleft}
x + \dfrac{b}{2a} &= \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}}\\
x + \dfrac{b}{2a} &= \pm \dfrac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} & \sqrt{4a^2} = |2a| = 2|a| = \pm 2a\\
x + \dfrac{b}{2a} &= \pm \dfrac{\sqrt{b^2 - 4ac}}{2a}\\
x &= -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} & \text{ Add these two fractions }\\
x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{array}\)
Sample Set B
Solve each of the following quadratic equations using the quadratic formula.
\(3x^2 + 5x + 2 = 0\)
1. Identify \(a, b\), and \(c\).
\(a = 3, b = 5, c = 2\)
2. Write the quadratic formula.
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
3. Substitute.
\(\begin{array}{flushleft}
x &= \dfrac{-5 \pm \sqrt{(5)^2 - 4(3)(2)}}{2(3)}\\
&= \dfrac{-5 \pm \sqrt{25 - 24}}{6}\\
&= \dfrac{-5 \pm \sqrt{1}}{6}\\
&= \dfrac{-5 + 1}{6} & -5 + 1 = -4 \text{ and } -5 - 1 = -6
&= \dfrac{-4}{6}, \dfrac{-6}{6}\\
x & \dfrac{-2}{3}, -1
\end{array}\)
Note: Since these roots are rational numbers, this equation could have been solved by factoring.
\(12x^2 - 2x - 1 = 0\)
1. Identify \(a, b\), and \(c\).
\(a = 12, b = -2, c = -1\)
2. Write the quadratic formula.
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
3. Substitute.
\(\begin{array}{flushleft}
x &= \dfrac{(-2) \pm \sqrt{(-2)^2 - 4(12)(-1)}}{2(12)}\\
&= \dfrac{2 \pm \sqrt{4 + 48}}{24} & \text{ Simplify }\\
&= \dfrac{2 \pm \sqrt{52}}{24} & \text{ Simplify }\\
&= \dfrac{2 \pm \sqrt{4 \cdot 13}}{24} & \text{ Simplify }\\
&= \dfrac{2 \pm 2\sqrt{13}}{24} & \text{ Reduce. Factor } 2 \text{ from the terms of the numerator.}\\
&= \dfrac{2(1 \pm \sqrt{13})}{24}\\
x &= \dfrac{1 \pm \sqrt{13}}{12}
\end{array}\)
\(2y^2 + 3 = 0\)
1. Identify \(a, b\), and \(c\).
\(a = 2, b = 0, c = 3\)
2. Write the quadratic formula.
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
3. Substitute.
\(\begin{array}{flushleft}
x &= \dfrac{-0 \pm \sqrt{0^2 - 4(2)(3)}}{2(2)}\\
x &= \dfrac{0 \pm \sqrt{-24}}{4}
\end{array}\)
This equation has no real number solution since we have obtained a negative number under the radical sign.
\(-8x^2 + 11x = 0\)
1. Identify \(a, b\), and \(c\).
\(a = -8, b = 11, c = 0\)
2. Write the quadratic formula.
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
3. Substitute.
\(\begin{array}{flushleft}
x &= \dfrac{-11 \pm \sqrt{11^2 - 4(-8)(0)}}{2(-8)}\\
&= \dfrac{-11 \pm \sqrt{121 - 0}}{-16} & \text{ Simplify }\\
&= \dfrac{-11 \pm \sqrt{121}}{-16} & \text{ Simplify }\\
&= \dfrac{-11 \pm 11}{-16}
x &= 0, \dfrac{11}{8}
\end{array}\)
\((3x + 1)(x - 4) = x^2 + x - 2\)
1. Write the equation in standard form.
\(\begin{array}{flushleft}
3x^2 - 11x - 4 &= x^2 + x - 2\\
2x^2 - 12x - 2 &= 0\\
x^2 - 6x - 1 &= 0
\end{array}\)
2. Identify \(a, b\), and \(c\).
\(a = 1, b = -6, c = -1\)
3. Write the quadratic formula.
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
4. Substitute.
\(\begin{array}{flushleft}
x &= \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2(1)}\\
&= \dfrac{6 \pm \sqrt{36 + 4}}{2}\\
&= \dfrac{6 \pm \sqrt{40}}{2}\\
&= \dfrac{6 \pm \sqrt{4 \cdot 10}}{2}\\
&= \dfrac{6 \pm 2\sqrt{10}}{2}\\
&= \dfrac{2(3 \pm \sqrt{10}}{2}
\end{array}\)
\(x = 3 \pm \sqrt{10}\)
Practice Set B
Solve each of the following quadratic equations using the quadratic formula.
\(2x^2 + 3x - 7 = 0\)
- Answer
-
\(x = \dfrac{-3 \pm \sqrt{65}}{4}\)
\(5a^2 - 2a - 1 = 0\)
- Answer
-
\(a = \dfrac{1 \pm \sqrt{6}}{5}\)
\(6y^2 + 5 = 0\)
- Answer
-
no real number solution
\(-3m^2 + 2m = 0\)
- Answer
-
Add texts here. Do not delete this text first.
Exercises
For the following problems, solve the equations using the quadratic formula.
\(x^2 - 2x - 3 = 0\)
- Answer
-
\(x=3, −1\)
\(x^2 + 5x + 6 = 0\)
\(y^2 - 5y + 4 = 0\)
- Answer
-
\(y=1, 4\)
\(a^2 + 4a - 21 = 0\)
\(a^2 + 12a + 20 = 0\)
- Answer
-
\(a=−2, −10\)
\(b^2 - 4b + 4 = 0\)
\(b^2 + 4b + 4 = 0\)
- Answer
-
\(b=−2\)
\(x^2 + 10x + 25 = 0\)
\(2x^2 - 5x - 3 = 0\)
- Answer
-
\(x = 3, -\dfrac{1}{2}\)
\(6y^2 + y - 2 = 0\)
\(4x^2 - 2x - 1 = 0\)
- Answer
-
\(x = \dfrac{1 \pm \sqrt{5}}{4}\)
\(3y^2 + 2y - 1 = 0\)
\(5a^2 - 2a - 3 = 0\)
- Answer
-
\(a = 1, -\dfrac{3}{5}\)
\(x^2 - 3x + 1 = 0\)
\(x^2 - 5x - 4 = 0\)
- Answer
-
\(x = \dfrac{5 \pm \sqrt{41}}{2}\)
\((x+2)(x−1)=1\)
\((a+4)(a−5)=2\)
- Answer
-
\(a = \dfrac{1 \pm \sqrt{89}}{2}\)
\((x−3)(x+3)=7\)
\((b−4)(b+4)=9\)
- Answer
-
\(b = \pm 5\)
\(x^2 + 8x = 2\)
\(y^2 = -5y + 4\)
- Answer
-
\(y = \dfrac{-5 \pm \sqrt{41}}{2}\)
\(x^2 = -3x + 7\)
\(x^2 = -2x - 1\)
- Answer
-
\(x=−1\)
\(x^2 + x + 1 = 0\)
\(a^2 + 3a - 4 = 0\)
- Answer
-
\(a=−4, 1\)
\(y^2 + y = -4\)
\(b^2 + 3b = -2\)
- Answer
-
\(b=−1, −2\)
\(x^2 + 6x + 8 = -x - 2\)
\(x^2 + 4x = 2x - 5\)
- Answer
-
No real number solution.
\(6b^2 + 5b - 4 = b^2 + b + 1\)
\(4a^2 + 7a - 2 = -2a + a\)
- Answer
-
\(\dfrac{-2 \pm \sqrt{6}}{2}\)
\((2x + 5)(x - 4) = x^2 -x + 2\)
\((x-4)^2 = 3\)
- Answer
-
\(x = 4 \pm \sqrt{3}\)
\((b - 6)^2 = 8\)
- Answer
-
\(b = 6 \pm 2\sqrt{2}\)
\((3-x)^2 = 6\)
\(3(x^2 + 1) = 2(x+7)\)
- Answer
-
\(x = \dfrac{1 \pm \sqrt{34}}{3}\)
\(2(y^2 - 3) = -3(y - 1)\)
\((x + 2)^2 = 4\)
\(-4(a^2 + 2) + 3 = 5\)
- Answer
-
No real number solution
\(-(x^2 + 3x - 1) = 2\)
Exercises For Review
Simplify \((\dfrac{x^8y^7z^5}{x^4y^6z^2})^2\)
- Answer
-
\(x^8y^2z^6\)
Write \(4a^{-6}b^2c^3a^5b^{-3}\) so that only positive exponents appear
Find the product: \((2y + 7)(3y - 1)\)
- Answer
-
\(6y^2 + 19y - 7\)
Simplify: \(\sqrt{80} - \sqrt{45}\)
Solve \(x^2 - 4x - 12 = 0\) by completing the square.
- Answer
-
\(x=−2, 6\)