Problem 60 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation using any convenient method. \((x-2)^{2}-9=0\)

Step-by-Step Solution

Answer

The roots of the given quadratic equation \((x-2)^{2}-9=0\) are \(x=5\) and \(x=-1\).

Step 1: Expression simplification.

The given equation is \((x-2)^{2}-9=0\). To begin with, expand the equation by squaring \(x-2\) . This gives us \(x^{2}-4x+4-9=0\). Simplifying further will result in \(x^{2}-4x-5=0\) which is more straightforward to solve.

Step 2: Factoring the equation.

Factor the equation \(x^{2}-4x-5=0\) . This is done by splitting the 'b' term in a standard form quadratic equation. The equation is factored to \((x-5)(x+1)=0\) .

Step 3: Finding the roots.

Now, to find the roots of the equation, set each factor equal to zero and solve for 'x'. Thus \(x-5=0 => x=5\) and \(x+1=0 => x=-1\). Thus the roots of the given equation are \(x=5 and x=-1\).

Key Concepts

Factoring Quadratics

Factoring quadratics is the process of breaking down a quadratic equation into a product of simpler binomial expressions. The general form of a quadratic equation is \[ ax^2 + bx + c = 0 \]. To factor a quadratic, you'll look for two numbers that both add up to \( b \) (the coefficient of the linear term) and multiply to \( ac \) (the constant term).

• First, ensure the quadratic is in standard form.
• Find two numbers that multiply to \( c \) and add to \( b \).
• Rewrite the middle term using these numbers.
• Factor by grouping, if necessary, to simplify the expression.

In the equation \( x^2 - 4x - 5 = 0 \), the two numbers that multiply to -5 and add to -4 are -5 and 1. Thus, we can factor the equation into \((x-5)(x+1)=0\). This step is crucial as it sets the stage for solving the quadratic by finding its roots.

Roots of Quadratic Equations

The roots of a quadratic equation are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). Once the quadratic is factored, we find its roots by setting each factor equal to zero. This is because if the product of two expressions equals zero, at least one of the expressions must be zero.

• Take each factor separately.
• Set it equal to zero.
• Solve for \( x \) in each case.

For the equation \((x-5)(x+1)=0\), you set each factor to zero: \( x-5=0 \) and \( x+1=0 \). Solving these gives the roots \( x=5 \) and \( x=-1 \). These are the solutions for our original quadratic equation. Finding the roots not only gives us the specific solutions but also allows for understanding the x-intercepts if graphed.

Expanding Binomials

Expanding binomials involves applying the distributive property to the product of two binomials. The expression \((x-2)^2\) is expanded by multiplying \((x-2)\) by itself. This helps in simplifying complex quadratic equations into their standard form, making them easier to factor and solve.

• Use the formula \((a-b)^2 = a^2 - 2ab + b^2\).
• Substitute \(a = x\) and \(b = 2\).
• Calculate to obtain \(x^2 - 4x + 4\).

In our original equation, \((x-2)^2 - 9 = 0\), by expanding \((x-2)^2\) we got \((x^2 - 4x + 4)\), which simplified the problem to \(x^2 - 4x - 5 = 0\). This expansion step is vital as it makes the equation suitable for applying methods like factoring to find solutions.