Problem 50 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(-x=\sqrt{\frac{3 x+7}{4}}\)

Step-by-Step Solution

Answer

The solution is \( x = -1 \).

Step 1: Eliminate the Square Root

Square both sides to eliminate the square root. This means \( -x = \sqrt{\frac{3x + 7}{4}} \) becomes \( (-x)^2 = \left( \sqrt{\frac{3x + 7}{4}} \right)^2 \) or \( x^2 = \frac{3x + 7}{4} \).

Step 2: Clear the Fraction

Multiply both sides of the equation by 4 to eliminate the fraction. This results in \( 4x^2 = 3x + 7 \).

Step 3: Rearrange into Standard Quadratic Form

Subtract 3x and 7 from both sides to set the equation to zero: \( 4x^2 - 3x - 7 = 0 \).

Step 4: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4, b = -3, \) and \( c = -7 \). This gives us: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(4)(-7)}}{2(4)} = \frac{3 \pm \sqrt{9 + 112}}{8} = \frac{3 \pm \sqrt{121}}{8} \] \[ x = \frac{3 \pm 11}{8} \] \[ x = \frac{3 + 11}{8} = \frac{14}{8} = \frac{7}{4} \] and \[ x = \frac{3 - 11}{8} = -\frac{8}{8} = -1 \]. This provides the potential solutions: \( x = \frac{7}{4} \) and \( x = -1 \).

Step 5: Check the Solutions

Substitute \( x = \frac{7}{4} \) back into the original equation: \[ -\frac{7}{4} = \sqrt{\frac{3(\frac{7}{4}) + 7}{4}} \] \[ -\frac{7}{4} = \sqrt{\frac{\frac{21}{4} + 7}{4}} \] \[ -\frac{7}{4} = \sqrt{\frac{\frac{21}{4} + \frac{28}{4}}{4}} = \sqrt{\frac{\frac{49}{4}}{4}} = \sqrt{\frac{49}{16}} = \frac{7}{4} \] It's evident that \( -\frac{7}{4} eq \frac{7}{4}\), so \( x = \frac{7}{4} \) is not a valid solution.

Step 6: Check the Solution

Substitute \( x = -1 \) back into the original equation: \[ -(-1) = \sqrt{\frac{3(-1) + 7}{4}} \] \[ 1 = \sqrt{\frac{-3 + 7}{4}} \] \[ 1 = \sqrt{\frac{4}{4}} \] \[ 1 = \sqrt{1} \] \[ 1 = 1 \] This confirms that \( x = -1 \) is a valid solution.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are in the form \(ax^2 + bx + c = 0\). For any quadratic equation, you can find the solutions (or roots) using: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] . Here, 'a', 'b', and 'c' are coefficients from the equation. The term under the square root, \(b^2 - 4ac\), is known as the discriminant. If it's positive, you get two real solutions. If it's zero, there's exactly one real solution. And if it's negative, there are two complex solutions. Using the quadratic formula allows for efficient and straightforward solving of any standard quadratic equation.

Square Roots

Square roots are used to 'reverse' the squaring of a number. When solving an equation that involves a square root, like \( \sqrt{\frac{3x + 7}{4}} \), it helps to square both sides to eliminate the square root, making the equation easier to solve. For example, squaring \( \sqrt{a} = b \) results in \( a = b^2 \). This method was used in Step 1 of the given solution to simplify the problem. Removing the square root is crucial since it transforms a potentially tricky equation into a more familiar form that can be handled with standard algebraic methods.

Checking Solutions

Always check your solutions by substituting them back into the original equation. This confirms their validity and helps to avoid errors. For instance, after finding the potential solutions in the given problem, we substituted them back in to verify correctness. This step ensures that the solutions satisfy the original equation. In our example, one potential solution \(x = \frac{7}{4}\) did not satisfy the equation, while the other, \(x = -1\), did. Checking solutions prevents errors and validates the context of your answer.

Fraction Elimination

Eliminating fractions simplifies the equation, making it easier to solve. In the given exercise, we had to clear the fraction \( \frac{3x + 7}{4}\). By multiplying both sides by 4, we eliminated the denominator, transforming the equation into \(4x^2 = 3x + 7\). This step is crucial as it reduces complexity and allows the use of other algebraic methods to solve the problem. In equations involving fractions, always look for opportunities to clear the denominators early to facilitate simpler manipulations and solutions.