10.4: Solving Quadratic Equations Using the Method of Extraction of Roots
- Page ID
- 49404
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Method Of Extraction Of Roots
Quadratic equations of the form \(x^2 - K = 0\) can be solved by the method of extraction of roots by rewriting it in the form \(x^2 = K\).
To solve \(x^2 = K\), we are required to find some number, \(x\), that when squared produces \(K\). This number, \(x\), must be a square root of \(K\). If \(K\) is greater than zero, we know that it prossesses two square roots, \(\sqrt{K}\) and \(-\sqrt{K}\). We also know that
\((\sqrt{K})^2) = (\sqrt{K})(\sqrt{K}) = K\) and \((-\sqrt{K}) = (-\sqrt{K})(-\sqrt{K}) = K\)
We now have two replacements for \(x\) that produce true statements when substitued into the equation. Thus, \(x = \sqrt{K}\) and \(x = -\sqrt{K}\) are both solutions to \(x^2 = K\). We use the notation \(x = \pm \sqrt{K}\) to denote both the principal and secondary square roots.
The Nature of Solutions
For quadratic equations of the form \(x^2 = K\),
- If \(K\) is greater than or equal to zero, the solutions are \(\pm \sqrt{K}\).
- If \(K\) is negative, no real number solutions exist.
- If \(K\) is zero, the only solution is \(0\).
Sample Set A
Solve each of the following quadratic equations using the method of extraction of roots.
\(\begin{array}{flushleft}
x^2 - 49 &= 0 & \text{ Rewrite }\\
x^2 &= 49\\
x &= \pm \sqrt{49}\\
x &= \pm 7
\end{array}\)
Check:
\(\begin{array}{flushleft}
(7)^2 = 49 & \text{ Is this correct? } & (-7)^2 = 49 & \text{ Is this correct? }\\
49 = 49 & \text{ Yes, this is correct. } & 49 = 49 & \text{ Yes, this is correct. }
\end{array}\)
\(\begin{array}{flushleft}
25a^2 &= 36\\
a^2 &= \dfrac{36}{25}\\
a &= \pm \sqrt{\frac{36}{25}}\\
a &= \pm \dfrac{6}{5}
\end{array}\)
Check:
\(\begin{array}{flushleft}
25(\dfrac{6}{5})^2 &= 36 & \text{ Is this correct? } & 25(\dfrac{-6}{5})^2 &= 36 & \text{ Is this correct? }\\
25(\dfrac{36}{25})^2 &= 36 & \text{ Is this correct? } & 25(\dfrac{36}{25}) &= 36 & \text{ Is this correct? }\\
36 &= 36 & \text{Yes, this is correct. } & 36 &= 36 & \text{ Yes, this is correct. }
\end{array}\)
\(\begin{array}{flushleft}
4m^2 - 32 &= 0\\
4m^2 &= 32\\
m^2 &= \dfrac{32}{4}\\
m^2 &= 8\\
m &= \pm \sqrt{8}\\
m &= \pm 2\sqrt{2}
\end{array}\)
Check:
\(\begin{array}{flushleft}
4(2\sqrt{2})^2 &= 32 & \text{ Is this correct? } & 4(-2\sqrt{2})^2 &= 32 & \text{ Is this correct? }\\
4[2^2(\sqrt{2})^2] &= 32 & \text{ Is this correct? } & 4[(-2)^2(\sqrt{2})^2] &= 32 & \text{ Is this correct? }\\
4[4 \cdot 2] &= 32 & \text{ Is this correct? } & 4[4 \cdot 2] &= 32 & \text{ Is this correct? }\\
4 \cdot 8 &= 32 & \text{ Is this correct? } & 4 \cdot 8 &= 32 & \text{ Is this correct? }\\
32 &= 32 & \text{ Yes, this is correct. } & 32 &=32 & \text{ Yes, this is correct. }
\end{array}\)
Solve \(5x^2 - 15y^2z^7 = 0\) for \(x\).
\(\begin{array}{flushleft}
5x^2 &= 15y^2z^7 & \text{ Divide both sides by } 5\\
x^2 &= 3y^2z^7\\
x &= \pm \sqrt{3y^2z^7}\\
x &= \pm yz^3\sqrt{3z}
\end{array}\)
Calculator Problem:
Solve \(14a^2 - 235 = 0\). Round to the nearest hundredth.
\(\begin{array}{flushleft}
14a^2 - 235 &= 0 & \text{ Rewrite }\\
14a^2 &= 235 & \text{ Divide both sides by } 14\\
a^2 &= \dfrac{235}{14}
\end{array}\)
Rounding to the nearest hundredth produces \(4.10\). We must be sure to insert the \(\pm\) symbol.
\(a \approx \pm 4.10\).
\(\begin{array}{flushleft}
k^2 &= -64\\
k &= \pm \sqrt{-64}
\end{array}\)
The radicand is negative so no real number solutions exist
Practice Set A
Solve each of the following quadratic equations using the method of extraction of roots.
\(x^2 - 144 = 0\)
- Answer
-
\(x = \pm 12\)
\(9y^2 - 121 = 0\)
- Answer
-
\(y = \pm \dfrac{11}{3}\)
\(6a^2 = 108\)
- Answer
-
\(a = \pm 3\sqrt{2}\)
Solve \(4n^2 = 24m^2p^8\) for \(n\).
- Answer
-
\(n = \pm mp^4\sqrt{6}\)
Solve \(5p^2q^2 = 45p^2\) for \(q\)
- Answer
-
\(q = \pm 3\)
Solve \(16m^2 - 2206 = 0\). Round to the nearest hundredth.
- Answer
-
\(m = \pm 11.74\)
\(h^2 = -100\)
Sample Set B
Solve each of the following quadratic equations using the method of extraction of roots.
\(\begin{array}{flushleft}
(x+2)^2 &= 81\\
x + 2 &= \pm \sqrt{81}\\
x + 2 &= \pm 9 & \text{ Subtract } 2 \text{ from both sides.}\\
x &= -2 \pm 9\\
x &= -2 + 9 & \text{and} & x&= -2 - 9\\
x &= 7 & & x &= -11
\end{array}\)
\(\begin{array}{flushleft}
(a+3)^2 &= 5\\
a + 3 &= \pm \sqrt{5} & \text{ Subtract } 3 \text{ from both sides }\\
a &= -3 \pm \sqrt{5}
\end{array}\)
Practice Set B
Solve each of the following quadratic equations using the method of extraction of roots.
\((a + 6)^2 = 64\)
- Answer
-
\(a=2,−14\)
\((m - 4)^2 = 15\)
- Answer
-
\(m = 4 \pm \sqrt{15}\)
\((y-7)^2 = 49\)
- Answer
-
\(y=0, 14\)
\((k - 1)^2 = 12\)
- Answer
-
\(k = 1 \pm 2 \sqrt{3}\)
\((x - 11)^2 = 0\)
- Answer
-
\(x=11\)
Exercises
For the following problems, solve each of the quadratic equations using the method of extraction of roots.
\(x^2 = 36\)
- Answer
-
\(x = \pm 6\)
\(x^2 = 49\)
\(a^2 = 9\)
- Answer
-
\(a = \pm 3\)
\(a^2 = 4\)
\(b^2 = 1\)
- Answer
-
\(b = \pm 1\)
\(a^2 = 1\)
\(x^2 = 25\)
- Answer
-
\(x = \pm 5\)
\(x^2 = 81\)
\(a^2 = 5\)
- Answer
-
\(a = \pm \sqrt{5}\)
\(a^2 = 10\)
\(b^2 = 12\)
- Answer
-
\(b = \pm 2\sqrt{3}\)
\(b^2 = 6\)
\(y^2 = 3\)
- Answer
-
\(y = \pm \sqrt{3}\)
\(y^2 = 7\)
\(a^2 - 8 = 0\)
- Answer
-
\(a = \pm 2\sqrt{2}\)
\(a^2 - 3 = 0\)
\(a^2 - 5 = 0\)
- Answer
-
\(a = \pm \sqrt{5}\)
\(y^2 - 1 = 0\)
\(x^2 - 10 = 0\)
- Answer
-
\(x = \pm \sqrt{10}\)
\(x^2 - 11 = 0\)
\(3x^2 - 27 = 0\)
- Answer
-
\(x = \pm 3\)
\(5b^2 - 5 = 0\)
\(2x^2 = 50\)
- Answer
-
\(x = \pm 5\)
\(4a^2 = 40\)
\(2x^2 = 24\)
- Answer
-
\(x = \pm 2\sqrt{3}\)
For the following problems, solve for the indicated variable.
\(x^2 = 4a^2\), for \(x\)
\(x^2 = 9b^2\), for \(x\)
- Answer
-
\(x = \pm 3b\)
\(a^2 = 25c^2\), for \(a\)
\(k^2 = m^2n^2\), for \(k\).
- Answer
-
\(k = \pm mn\)
\(k^2 = p^2q^2r^2\), for \(k\)
\(2y^2 = 2a^2n^2\), for \(y\)
- Answer
-
\(y = \pm an\)
\(9y^2 = 27x^2z^4\), for \(y\).
\(x^2 - z^2 = 0\), for \(x\)
- Answer
-
\(x = \pm z\)
\(x^2 - z^2 = 0\), for \(z\)
\(5a^2 - 10b^2 = 0\), for \(a\)
- Answer
-
\(a = b\sqrt{2}, -b\sqrt{2}\)
For the following problems, solve each of the quadratic equations using the method of extraction of roots.
\((x-1)^2 = 4\)
\((x-2)^2 = 9\)
- Answer
-
\(x=5,−1\)
\((x-3)^2 = 25\)
\((a-5)^2 = 36\)
- Answer
-
\(x=11,−1\)
\((a + 3)^2 = 49\)
\((a + 9)^2 = 1\)
- Answer
-
\(a=−8 ,−10\)
\((a - 6)^2 = 3\)
\((x + 4)^2 = 5\)
- Answer
-
\(a = -4 \pm \sqrt{5}\)
\((b + 6)^2 = 7\)
\((x + 1)^2 = a\), for \(x\).
- Answer
-
\(x = -1 \pm \sqrt{a}\)
\((y + 5)^2 = b\), for \(y\)
\((y + 2)^2 = a^2\), for \(y\)
- Answer
-
\(y = -2 \pm a\)
\((x + 10)^2 = c^2\), for \(x\)
\((x - a)^2 = b^2\), for \(x\)
- Answer
-
\(x = a \pm b\)
\((x + c)^2 = a^2\), for \(x\)
Calculator Problems
For the following problems, round each result to the nearest hundredth.
\(8a^2 - 168 = 0\)
- Answer
-
\(a = \pm 4.58\)
\(6m^2 - 5 = 0\)
\(0.03y^2 = 1.6\)
- Answer
-
\(y = \pm 7.30\)
\(0.048x^2 = 2.01\)
\(1.001x^2 - 0.999 = 0\)
- Answer
-
\(x = \pm 1.00\)
Exercises For Review
Graph the linear inequality \(3(x + 2) < 2(3x + 4)\)
Solve the fractional equation: \(\dfrac{x-1}{x + 4} = \dfrac{x + 3}{x - 1}\)
- Answer
-
\(x = \dfrac{-11}{9}\)
Find the product: \(\sqrt{32x^3y^5} \sqrt{2x^3y^3}\)
Solve \(x^2 - 4x = 0\)
- Answer
-
\(x = 0, 4\)
Solve \(y^2 - 8y = -12\)