Problem 42 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(4 x=\sqrt{6 x+1}\)

Step-by-Step Solution

Answer

The solution to the equation is \( x = \frac{1}{2}\).

Step 1: Isolate the square root

The equation is given as i.e. \( \~4x = \sqrt{6x +1} \)

Step 2: 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: \( \16x^2 = 6x + 1 \)

Step 3: Form a quadratic equation

Move all terms to one side to form a quadratic equation: \( \16x^2 - 6x - 1 = 0\)

Step 4: 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 \[ x = \frac{6 - 10}{32} = \frac{-4}{32} = \frac{-1}{8} \]

Step 5: 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.

Key Concepts

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:

• \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(16)(-1)}}{2(16)}\)
• 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 \[ x = \frac{6 - 10}{32} = \frac{-4}{32} = \frac{-1}{8} \]

Understanding how to use and apply the quadratic formula is a valuable algebraic skill.

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.

This step is crucial in ensuring your final answers are correct and valid.

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.

By isolating the square root, we set up the equation for straightforward manipulation, such as squaring both sides to eliminate the square root.

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: \(16x^2 = 6x + 1\).

After squaring, the equation turns into a quadratic equation \(16x^2 - 6x - 1 = 0\), which can then be solved using the quadratic formula. It's essential to square both sides correctly and to not forget that new solutions might need to be checked for validity.