9.7: Square Root Equations with Applications
- Page ID
- 49399
Square Root Equations And Extraneous Solutions
Square Root Equation
A square root equation is an equation that contains a variable under a square root sign. The fact that \(\sqrt{x} \cdot \sqrt{x} = (\sqrt{x})^2 = x\) suggests that we can solve a square root equation by squaring both sides of the equation.
Extraneous Solutions
Squaring both sides of an equation can, however, introduce extraneous solutions. Consider the equation
\(x = -6\)
The solution is \(-6\). Square both sides.
\(x^2 = (-6)^2\)
\(x^2 = 36\)
This equation has two solutions, \(-6\) and \(+6\). The \(+6\) is an extraneous solution since it does not check in the original equation:
\(+6 \not = -6\)
Method For Solving Square Root Equations
- Isolate a radical. This means get a square root expression by itself on one side of the equal sign.
- Square both sides of the equation.
- Simplify the equation by combining like terms.
- Repeat step 1 if radicals are still present.
- Obtain potential solutions by solving the resulting non-square root equation.
- Check each potential solution by substitution into the original equation.
Sample Set A
Solve each square root equation.
\(\begin{array}{flushleft}
& \sqrt{x} &= 8 & \text{ The radical is isolated Square both sides. }\\
& (\sqrt{x})^2 &= 8^2\\
& x &= 64 & \text{ Check this potential solution }\\
\text{Check: } & \sqrt{64} &= 8 & \text{ Is this correct? }\\
& 8 &= 8 & \text{ Yes, this is correct. }\\
64 \text{ is the solution }
\end{array}\)
\(\begin{array}{flushleft}
& \sqrt{y-3} &= 4 & \text{ The radical is isolated. Square both sides. }\\
& \sqrt{y-3} &= 16 & \text{ Solve this nonradical equation }\\
\text{Check: } & \sqrt{19 - 3} &= \sqrt{16} & \text{ Is this correct? }\\
& \sqrt{16} &= 4 & \text{ Is this correct? }\\
& 4 &= 4 & \text{ Yes, this is correct. }\\
19 \text{ is the solution}
\end{array}\)
\(\begin{array}{flushleft}
& \sqrt{2m + 3} - \sqrt{m - 8} &= 0 & \text{ Isolate either radical }\\
& \sqrt{2m + 3} &= \sqrt{m + 8} & \text{ Square both sides. }\\
& 2m + 3 &= m-8 & \text{ Solve this nonradical equation }\\
& m &= -11 & \text{ Check this potential solution. }\\
\text{Check: } \sqrt{2(-11) + 3} - \sqrt{(-11) - 8} &= 0 & \text{ Is this correct? }\\
& \sqrt{-22 + 3} - \sqrt{-19} &= 0 & \text{ Is this correct? }
\end{array}\)
Since \(\sqrt{-19}\) is not a real number, the potential solution of \(m = -11\) does not check. This equation has no real solution.
\(\sqrt{4x - 5} = -6\). By inspection, this equation has no real solution.
The symbol, \(\sqrt{}\), signifies the positive square root and not the negative square root.
Practice Set A
Solve each square root equation.
\(\sqrt{y} = 14\)
- Answer
-
\(y = 196\)
\(\sqrt{a - 7} = 5\)
- Answer
-
\(a = 32\)
\(\sqrt{3a + 8} - \sqrt{2a + 5} = 0\)
- Answer
-
\(a = -3\) is extraneous, no real solution.
\(\sqrt{m - 4} = -11\)
- Answer
-
No real solution
Exercises
For the following problems, solve the square root equations.
\(\sqrt{x} = 5\)
- Answer
-
\(x = 25\)
\(\sqrt{y} = 7\)
\(\sqrt{a} = 10\)
- Answer
-
\(a = 100\)
\(\sqrt{c} = 12\)
\(\sqrt{x} = -3\)
- Answer
-
No solution
\(\sqrt{y} = -6\)
\(\sqrt{x} = 0\)
- Answer
-
\(x = 0\)
\(\sqrt{x} = 1\)
\(\sqrt{x + 3} = 3\)
- Answer
-
\(x = 6\)
\(\sqrt{y - 5} = 5\)
\(\sqrt{a + 2} = 6\)
- Answer
-
\(a = 34\)
\(\sqrt{y + 7} = 9\)
\(\sqrt{y - 4} - 4 = 0\)
- Answer
-
\(y = 20\)
\(\sqrt{x - 10} - 10 = 0\)
\(\sqrt{x - 16} = 0\)
- Answer
-
\(x = 16\)
\(\sqrt{y -25} = 0\)
\(\sqrt{6m - 4} = \sqrt{5m - 1}\)
- Answer
-
\(m = 3\)
\(\sqrt{5x + 6} = \sqrt{3x + 7}\)
\(\sqrt{7a + 6} = \sqrt{3a - 18}\)
- Answer
-
No solution
\(\sqrt{4x + 3} = \sqrt{x - 9}\)
\(\sqrt{10a - 7} - \sqrt{2a + 9} = 0\)
- Answer
-
\(a = 2\)
\(\sqrt{12k - 5} - \sqrt{9k + 10} = 0\)
\(\sqrt{x - 6} - \sqrt{3x - 8} = 0\)
- Answer
-
No solution
\(\sqrt{4a - 5} - \sqrt{7a - 20} = 0\)
\(\sqrt{2m - 6} = \sqrt{m - 2}\)
- Answer
-
\(m = 4\)
\(\sqrt{6r - 11} = \sqrt{5r + 3}\)
\(\sqrt{3x + 1} = \sqrt{2x - 6}\)
- Answer
-
No solution
\(\sqrt{x - 7} - \sqrt{5x + 1} = 0\)
\(\sqrt{2a + 9} - \sqrt{a - 4} = 0\)
- Answer
-
No solution
At a certain electronics company, the daily output \(Q\) is related to the number of people \(A\) on the assembly line by \(Q = 400 + 10\sqrt{A + 125}\)
a) Determine the daily output if there are \(44\) people on the assembly line.
b) Determine how many people are needed on the assembly line if the daily output is to be \(520\)
At a store, the daily number of sales \(S\) is approximately related to the number of employees \(E\) by \(S = 100 + 15\sqrt{E + 6}\)
a) Determine the approximate number of sales if there are \(19\) employees.
b) Determine the number of employees the store would need to produce \(310\) sales.
- Answer
-
a) \(S = 175\)
b) \(E = 190\)
The resonance frequency \(f\) in an electronic circuit containing inductance \(L\) and capacitance \(C\) in series is given by:
\(f = \dfrac{1}{2 \pi \sqrt{LC}}\)
a) Determine the resonance frequency in an electronic circuit if the inductance is \(4\) and the capacitance is \(0.0001\). Use \(\pi = 3.14\)
b) Determine the inductance in an electric circuit if the resonance frequency is \(7.12\) and the capacitance is \(0.0001\). Use \(\pi = 3.14\).
If two magnetic poles of strength \(m\) and \(m'\) units are at a distance \(r\) centimeters (cm) apart, the force \(F\) of repulsion in the air between them is given by:
\(F = \dfrac{mm'}{r^2}\):
a) Determine the force of repulsion if two magnetic poles of strengths \(20\) and \(40\) are \(5\) cm apart in the air.
b) Determine how far apart are two magnetic poles of strengths \(30\) and \(40\) units if the force of repulsion in the air between them is \(0.0001\).
- Answer
-
a) \(F = 32\)
b) \(r = 8\)cm.
The velocity \(V\) in feet per second of outflow of a liquid from an orifice is given by \(V = 8\sqrt{h}\), where \(h\) is the height in feet of the liquid above the opening.
a) Determine the velocity of outflow of a liquid from an orifice that is \(9\) feet below the top surface of a liquid (\(V\) is in feet/sec.
b) Determine how high a liquid is above an orifice if the velocity of outflow is \(81\) feet/second.
The period \(T\) in seconds of a simple pendulum of length \(L\) in feet is given by \(T = 2 \pi \sqrt{\dfrac{L}{32}}\)
a) Determine the period of a simple pendulum that is \(2\) feet long. Use \(\pi = 3.14\).
b) Determine the length in feet of a simple pendulum whose period if \(10.8772\) seconds. Use \(\pi = 3.14\).
- Answer
-
a) \(T = 1.57\)sec
b) \(L = 95.99\)cm
The kinetic energy \(KE\) in foot pounds of a body of mass \(m\) in slugs moving with a velocity \(v\) in feet/sec is given by \(KE = \dfrac{1}{2}mv^2\).
a) Determine the kinetic energy of a \(2\)-slug body moving with a velocity of \(4\) ft/sec.
b) Determine the velocity in feet/sec of a \(4\)-slug body if its kinetic energy is \(50\) foot-pounds.
Exercises For Review
Write \(\dfrac{x^{10}y^3(x+7)^4}{x^{-2}y^3(x+7)^{-1}}\) so that only positive exponents appear.
- Answer
-
\(x^{12}(x+7)^5\)
Classify \(x+4 = x+7\) as an identity, a contradiction, or a conditional equation.
Supply the missing words. In the coordinate plane, lines with _____ slope rise and lines with _____ slope fall.
- Answer
-
positive, negative
Simplify \(\sqrt{(x+3)^4(x-2)^6}\)
Simplify \((3 + \sqrt{5})(4 - \sqrt{5})\)
- Answer
-
\(7 + \sqrt{5}\)