Problem 90 - Algebra for College Students

Question

Solve each equation. $$(5 x+21)^{\frac{1}{2}}=x+3$$

Step-by-Step Solution

Answer

The solution is \(x = 3\).

Step 1: Understand the Problem

We have the equation \((5x + 21)^{\frac{1}{2}} = x + 3\). The left side of the equation is a square root expression, and our goal is to isolate \(x\) and solve the equation.

Step 2: Eliminate the Square Root

To remove the square root, square both sides of the equation. Squaring gives us \((5x + 21) = (x + 3)^2\).

Step 3: Expand the Right Side

Expand \((x + 3)^2\) to get \(x^2 + 6x + 9\). Our equation now becomes \(5x + 21 = x^2 + 6x + 9\).

Step 4: Rearrange the Equation

Subtract \(5x\) and 21 from both sides to set the equation to zero, getting \(x^2 + 6x + 9 - 5x - 21 = 0\). This simplifies to \(x^2 + x - 12 = 0\).

Step 5: Solve the Quadratic Equation

We solve \(x^2 + x - 12 = 0\) using factoring, where possible. The factors of \(-12\) that add up to \(1\) are \(4\) and \(-3\). Thus, the equation factors to \((x + 4)(x - 3) = 0\).

Step 6: Find the Solutions

Set each factor equal to zero: \(x + 4 = 0\) gives \(x = -4\), and \(x - 3 = 0\) gives \(x = 3\). These are our solutions.

Step 7: Verify the Solutions

Substitute \(x = -4\) back into the original equation: \((5(-4) + 21)^{\frac{1}{2}} = \sqrt{1} = 1\) which does not equal \(-1\). So \(x = -4\) is invalid. Now substitute \(x = 3\): \((5(3) + 21)^{\frac{1}{2}} = \sqrt{36} = 6\) which equals \(6\). \(x = 3\) is valid.

Key Concepts

Quadratic Equations

Quadratic equations are equations where the highest exponent of the variable is 2. They're usually given in the form \( ax^2 + bx + c = 0 \). In our exercise, we derived the quadratic equation \( x^2 + x - 12 = 0 \). To recognize a quadratic equation, look for the squared term, which is what distinguishes it from linear equations.

Solving quadratic equations can be done by several methods, such as:

• Factoring: This involves writing the equation as a product of two binomials.
• Quadratic Formula: Provides solutions using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the Square: Rewriting the equation as a perfect square to easily solve the variable.

Each method has its own advantages. In our exercise, we used factoring because the equation can be easily expressed as a product of binomials.

Factoring

Factoring is a method used to solve quadratic equations by expressing them as the product of simpler expressions, also known as binomials. When we say to factor, we're looking for two numbers that multiply to the constant term and add to the coefficient of the linear term. In the case of \( x^2 + x - 12 \):

• These numbers are \( 4 \) and \( -3 \) because \( 4 \times -3 = -12 \) and \( 4 + (-3) = 1 \).

So, the equation can be factored into \( (x + 4)(x - 3) = 0 \). This method is quite efficient when these numbers exist neatly. If factoring is complex or not possible, other methods like the quadratic formula might be more suitable.

Once factored, setting each binomial to zero gives potential solutions for \( x \). The logic is that if a product equals zero, one or both of the factors must equal zero. Solving these gives the solutions to the quadratic equation.

Square Roots

The concept of square roots is vital in solving equations involving radicals. In our exercise, the equation \( (5x + 21)^{\frac{1}{2}} = x + 3 \) begins with a square root, \( (5x + 21)^{\frac{1}{2}} \). To simplify solving, we eliminate the square root by squaring both sides.

When you square both sides, be cautious as it can introduce extraneous solutions that don't satisfy the original equation. That's why verifying solutions is crucial. In the exercise:

• Squaring gave us \( 5x + 21 = (x + 3)^2 \).
• It's important to ensure each derived solution fits into the initial equation.

The square root is fundamental when checking answers. Since squaring can disguise mistakes, plugging solutions back to the original equation helps verify they are correct. Square roots are thus not only operators but also tools to check validity in mathematical solutions.