Problem 76 - Elementary Algebra

Question

Problem: Solve: \(x^{2}-2 x-35=0\) Incorrect Answer: \(b=-2 ;\left(\frac{1}{2}(-2)\right)^{2}=1\) \(x^{2}-2 x-35=0\) \(\frac{+35+35}{x^{2}-2 x+0=35}\) \begin{tabular}{r} \(+1+1\) \\ \hline\(x^{2}-2 x+1=36\) \end{tabular} \((x+1)(x+1)=36\) \((x+1)^{2}=36\) \(x+1=\sqrt{36}\) or \(x+1=-\sqrt{36}\) \(x+1=6\) or \(x+1=-6\) \(\frac{-1-1}{x+0=5} \quad\) or \(\quad \frac{-1}{x+0=-7}\) \(x=5 \quad\) or \(\quad x=-7\)

Step-by-Step Solution

Answer

The solutions are \( x = -5 \) and \( x = 7 \)

Step 1: Identify and rewrite the quadratic

The given equation is a quadratic equation of the form: \(x^2 - 2x - 35 = 0\)

Step 2: Factorize the quadratic equation

To solve the equation, look for two numbers that multiply to -35 and add up to -2. These numbers are 5 and -7. The equation can be factorized as: \( (x + 5)(x - 7) = 0 \)

Step 3: Apply the zero-product property

The zero-product property states that if \( (a)(b) = 0 \), then either \( a = 0 \) or \( b = 0 \). Therefore, set each factor equal to zero: \( x + 5 = 0 \) \( x - 7 = 0 \)

Step 4: Solve for x

Solve each equation for \( x\): \( x + 5 = 0 \implies x = -5 \) \( x - 7 = 0 \implies x = 7 \)

Step 5: Verify the solutions

Substitute \( x = -5 \) and \( x = 7 \) back into the original equation to ensure they satisfy it. Both values should make the original equation true.

Key Concepts

Factoring Quadratics

When solving quadratic equations, one common method is factoring. Factoring quadratics involves breaking down a quadratic expression into two simpler binomials that, when multiplied, give back the original quadratic expression.
Consider the quadratic equation: \(x^2 - 2x - 35 = 0\).
We need to find two numbers that multiply to \(-35\) and add up to \(-2\). In this case, those numbers are \(5\) and \(-7\).
Hence, we can rewrite the expression as \((x + 5)(x - 7) = 0\).

Factoring quadratics not only simplifies the solving process but also provides insight into the roots of the equation. It's a useful technique for polynomials where factors are easily identifiable.

Zero-Product Property

The zero-product property is a fundamental principle in algebra. It states that if the product of two factors is zero, then at least one of the factors must be zero. This property is particularly useful when solving factored quadratic equations.

Let's apply this property to our factored equation:\((x + 5)(x - 7) = 0\). According to the zero-product property, set each factor to zero:
- \(x + 5 = 0\)
- \(x - 7 = 0\)

This means that the solutions to the equation can be found by solving these simpler equations. This technique turns a seemingly complex quadratic problem into two straightforward linear equations.

Quadratic Solutions

Once we have factored the quadratic and applied the zero-product property, the next step is to find the solutions. By solving each resulting linear equation, we get our quadratic solutions.

For the equations we found:
- \(x + 5 = 0\)
- \(x - 7 = 0\)
Solving these gives:
- \(x = -5\)
- \(x = 7\)

These values are the solutions to the original quadratic equation. To confirm they are correct, substitute \(x = -5\) and \(x = 7\) back into the original equation \(x^2 - 2x - 35 = 0\) and verify that both make the equation true.

Understanding quadratic solutions is essential as it ensures that you have correctly factored and applied algebraic principles to find the roots of quadratic equations. This method of solving quadratics by factoring and applying the zero-product property provides a clear and systematic route to finding the solutions.