Problem 93 - Algebra for College Students

Question

Suppose that your friend solved the equation \((x+3)^{2}=\) 25 as follows: $$ \begin{aligned} (x+3)^{2} &=25 \\ x^{2}+6 x+9 &=25 \\ x^{2}+6 x-16 &=0 \end{aligned} $$ $$ \begin{aligned} (x+8)(x-2) &=0 & & & \\ x+8 &=0 \quad \text { or } & & x-2 &=0 \\ x &=-8 & \text { or } & x &=2 \end{aligned} $$ Is this a correct approach to the problem? Would you offer any suggestion about an easier approach to the problem?

Step-by-Step Solution

Answer

Your friend's approach is correct, but only \(x = 2\) satisfies the original equation.

Step 1: Expand the Equation

Your friend started by expanding the equation \((x+3)^2 = 25\). This expanded to:\[ x^2 + 6x + 9 = 25 \]which is mathematically correct. Then they rearranged it to form a quadratic equation:\[ x^2 + 6x + 9 - 25 = 0 \] evolving to:\[ x^2 + 6x - 16 = 0 \].

Step 2: Factor the Quadratic

Next, your friend factored the quadratic equation \(x^2 + 6x - 16 = 0\). They correctly factored it into two binomials:\[(x + 8)(x - 2) = 0 \]This step is also correct.

Step 3: Solve the Factored Equation

From the factored equation, they found the solutions by setting each factor equal to zero:\[ x + 8 = 0 \quad \text{or} \quad x - 2 = 0 \]Solving these gives:\[ x = -8 \quad \text{or} \quad x = 2 \].

Step 4: Verify the Solutions

To verify, substitute \(x = -8\) and \(x = 2\) back into the original equation. Only \(x = 2\) satisfies the original equation because \((2+3)^2 = 25\) while \((-8+3)^2 eq 25\).

Step 5: Alternative Easier Approach

An alternative, simpler method is to take the square root of both sides directly from \((x+3)^2 = 25\):\[ x+3 = \pm 5 \]which gives:\[ x+3 = 5 \quad \text{or} \quad x+3 = -5 \]Solving these linear equations:\[ x = 2 \quad \text{or} \quad x = -8 \]. However, upon final confirmation, only \(x = 2\) is valid, checking the solutions as discussed in the prior step.

Key Concepts

Factoring Quadratics

Factoring quadratics is an essential technique in solving quadratic equations, particularly when the equation is in the standard form \( ax^2 + bx + c = 0 \). The goal is to express the quadratic expression as a product of two binomials. For example, if we have the equation \( x^2 + 6x - 16 = 0 \), the solution involves finding two numbers that multiply to (-16) (the constant term, \(c\)) and also sum up to \(6\) (the coefficient of \(x\), \(b\)).

In this case, those two numbers are \(8\) and \(-2\). Therefore, the quadratic can be factored as \( (x + 8)(x - 2) = 0 \). Factoring not only simplifies the equation, making it easier to solve, but also provides a clear structure to find the solutions.

• Look for two numbers that multiply to \(ac\) and add to \(b\).
• Use these numbers to split the middle term and factor in pairs.

Solving Equations

Once the quadratic equation is factored, solving it becomes much more straightforward. You rely on the "zero product property," which states that if a product equals zero, at least one of the factors must be zero.

From the example given, \( (x+8)(x-2) = 0 \) implies \( x+8 = 0 \) or \( x-2 = 0 \). Solving these simple linear equations provides your potential solutions: \( x = -8 \) and \( x = 2 \). It's crucial to inspect each solution's validity against the original equation.

Verifying Solutions

Verification is a vital step in solving equations to ensure the solutions satisfy the original equation. In this process, you substitute each solution back into the beginning equation to check for accuracy.

For the quadratic equation \( (x+3)^2 = 25 \), substitute \( x = 2 \): \( (2+3)^2 = 5^2 = 25 \) hence it is correct. However, for \( x = -8 \), \( (-8+3)^2 = (-5)^2 = 25 \) seems correct. But the expanded equation shows \( x = -8 \) does not truly satisfy \( x^2 + 6x - 16 = 0 \). Therefore, only \( x = 2 \) is valid after re-evaluation.

Alternative Methods

Quadratic equations can often be solved using various methods, such as factoring, completing the square, or using the quadratic formula, depending on the structure of the equation. In the original problem, another efficient method is recognizing that since \((x+3)^2 = 25\), we can take the square root of both sides directly.

By taking the square roots, we obtain two scenarios: \(x+3 = 5\) or \(x+3 = -5\). Solving these linear equations provides \(x = 2\) and \(x = -8\). It's vital to evaluate both solutions to make sure they satisfy the complete equation, reaffirming that sometimes a quicker method might check the solution faster and with less chance of error.