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Magazine Chapter 11: Problem 42
Solve each equation. Check the solutions. \(4 x=\sqrt{6 x+1}\)
Short Answer
Expert verified
The solution to the equation is \( x = \frac{1}{2}\).
Step by step solution
01
Isolate the square root
The equation is given as i.e. \( \~4x = \sqrt{6x +1} \)
02
Square both sides
Square both sides of the equation to eliminate the square root. This becomes: \( (4x)^2 = (\sqrt{6x +1})^2 \) Simplifying both sides, we get: \( \16x^2 = 6x + 1 \)
03
Form a quadratic equation
Move all terms to one side to form a quadratic equation: \( \16x^2 - 6x - 1 = 0\)
04
Use the quadratic formula
To solve the quadratic equation, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 16 \), \( b = -6 \), and \( c = -1 \) This gives: \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(16)(-1)}}{2(16)} \) \(x = \frac{6 \pm \sqrt{36 + 64}}{32} \) \( x = \frac{6 \pm \sqrt{100}}{32} \) \( x = \frac{6 \pm 10}{32} \) Hence, we have two solutions: \[ x = \frac{6 + 10}{32} = \frac{16}{32} = \frac{1}{2} \] and \[ x = \frac{6 - 10}{32} = \frac{-4}{32} = \frac{-1}{8} \]
05
Check the solutions
Substitute the solutions back into the original equation to check if they are correct. For \( x = \frac{1}{2} \): \( 4(\frac{1}{2}) = \sqrt{6(\frac{1}{2}) + 1} \) \( 2 = \sqrt{3 + 1} \) \( 2 = \sqrt{4} \) \( 2 = 2 \) Thus, \( \frac{1}{2} \) is a solution. For \( x = -\frac{1}{8} \): \( 4(-\frac{1}{8}) = \sqrt{6(-\frac{1}{8}) + 1} \) \( -\frac{1}{2} = \sqrt{-\frac{3}{4} + 1} \) \( -\frac{1}{2} = \sqrt{\frac{1}{4}} \) \( -\frac{1}{2} eq \frac{1}{2} \) Thus, \( -\frac{1}{8} \) is not a solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic formula
The quadratic formula is a tool used to find the solutions of a quadratic equation, which is an equation of the form \(ax^2 + bx + c = 0\). The formula is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\). In this problem, the equation after squaring both sides and simplifying becomes \(16x^2 - 6x - 1 = 0\). Here, \(a = 16\), \(b = -6\), and \(c = -1\).
The quadratic formula then gives us two potential solutions for \(x\), which we can calculate by plugging in these values:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\). In this problem, the equation after squaring both sides and simplifying becomes \(16x^2 - 6x - 1 = 0\). Here, \(a = 16\), \(b = -6\), and \(c = -1\).
The quadratic formula then gives us two potential solutions for \(x\), which we can calculate by plugging in these values:
- \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(16)(-1)}}{2(16)}\)
- Simplifying under the square root: \(x = \frac{6 \pm \sqrt{36 + 64}}{32}\)
- Further simplifying: \(x = \frac{6 \pm 10}{32}\)
- That results in two possible values: \[ x = \frac{6 + 10}{32} = \frac{16}{32} = \frac{1}{2} \] and \[ x = \frac{6 - 10}{32} = \frac{-4}{32} = \frac{-1}{8} \]
checking solutions
After finding the potential solutions to a quadratic equation, it's important to check if they actually satisfy the original equation. This can prevent errors that might occur from extraneous solutions.
- For the solution \(x = \frac{1}{2}\), substitute it back into the original equation \(4x = \sqrt{6x + 1}\):
\(4(\frac{1}{2}) = \sqrt{6(\frac{1}{2}) + 1}\)
\(2 = \sqrt{3 + 1}\)
\(2 = \sqrt{4}\)
\(2 = 2\), which is true, so \(\frac{1}{2}\) is a valid solution. - For the solution \(x = -\frac{1}{8}\), substitute it back into the original equation:
\(4(-\frac{1}{8}) = \sqrt{6(-\frac{1}{8}) + 1}\)
\(-\frac{1}{2} = \sqrt{-\frac{3}{4} + 1}\)
\(-\frac{1}{2} = \sqrt{\frac{1}{4}}\)
\(-\frac{1}{2} eq \frac{1}{2}\), which is false. Hence, \(-\frac{1}{8}\) is not a valid solution.
isolate square root
When dealing with an equation involving a square root, it's essential to isolate the square root term before proceeding with any further operations. This simplifies the equation and makes it easier to solve.
- In the given problem, the original equation is \(4x = \sqrt{6x + 1}\).
- Here, the square root term \(\sqrt{6x + 1}\) is already isolated on one side of the equation.
squaring both sides
Squaring both sides of an equation is a common technique used to eliminate a square root. This step transforms the equation into a polynomial equation, which is easier to solve using algebraic methods.
- In our example, start with the isolated square root equation \(4x = \sqrt{6x + 1}\).
- Square both sides to remove the square root:
\((4x)^2 = (\sqrt{6x + 1})^2\) - This simplifies to: \(16x^2 = 6x + 1\).
