Problem 65 - Introductory Algebra for College Students

Question

Solve each equation and check your solutions. $$(x-2)^{2}-5(x-2)+6=0$$

Step-by-Step Solution

Answer

The solutions to the equation \((x-2)^{2}-5(x-2)+6=0\) are \(x_1=5\) and \(x_2=4\).

Step 1: Rewrite the Equation

Rewrite the equation to express \((x-2)^{2}-5(x-2)+6=0\) as \(y^{2}-5y+6=0\) with \(y=(x-2)\).

Step 2: Solve the Quadratic Equation

Now find the roots of the equation \(y^{2}-5y+6=0\) using the quadratic formula \(y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here a=1, b=-5 and c=6.

Step 3: Calculate the Discriminant

The discriminant, \(D = b^2-4ac\). Substitute a=1, b=-5 and c=6 to get \(D = (-5)^{2}-4*1*6 = 1\).

Step 4: Calculate the Roots

Since the discriminant > 0, we have two distinct real roots. The roots are given by \(y=\frac{-(-5)\pm\sqrt{1}}{2*1}\) or \(y_1=3\), \(y_2=2\). Substitute y with \(x-2\) to get \(x_1=5\), \(x_2=4\).

Step 5: Check the Solutions

Substitute \(x_1=5\) and \(x_2=4\) in the original equation \((x-2)^{2}-5(x-2)+6=0\) to verify if they are correct.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants. To use the formula, identify the coefficients \(a\), \(b\), and \(c\) from the equation, then plug them into \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula will provide you with the roots of the equation, which are the values of x when the equation equals zero.

When you have an equation like \(y^2 - 5y + 6 = 0\), applying the quadratic formula is straightforward. Simply identify \(a=1\), \(b=-5\), and \(c=6\), then plug these into the formula to find the roots. This process leads to two solutions for the variable y, and subsequently, the variable x after doing the substitution if the equation was initially transformed.

Factoring Quadratics

Factoring quadratics is another technique for finding the roots of a quadratic equation, particularly when the equation can be easily expressed as a product of two binomials. The general form of a quadratic equation \(ax^2 + bx + c = 0\) can sometimes be factored into \(\left(x - n\right)\left(x - m\right)=0\) where \(n\) and \(m\) are the roots of the equation. The factorized form makes it easy to see that when \(x=n\) or \(x=m\), the equation holds true, since either product becomes zero, resulting in the entire expression equating to zero.

For instance, in the step-by-step solution, the equation \(y^2 - 5y + 6 = 0\) is factorable into \(\left(y - 3\right)\left(y - 2\right)=0\), revealing the roots directly as \(y=3\) and \(y=2\). When possible, factoring is often quicker than using the quadratic formula, but it requires that the quadratic be factorable, which is not always the case.

Discriminant in Quadratics

Understanding the discriminant in quadratics gives you insight into the nature of the roots you'll find before you even calculate them. The discriminant is the part of the quadratic formula under the square root, \(b^2 - 4ac\), and it determines the type and number of roots of the quadratic equation. If the discriminant is positive, there are two distinct real roots; if it's zero, there is one real double root (the graph of the function touches the x-axis at one point); and if it's negative, there are two complex roots (and thus the function does not intersect the x-axis).

In the worked exercise, the discriminant is calculated as \(D = (-5)^2 - 4\times1\times6 = 1\), which is positive, indicating that two distinct real roots exist. This numerical value of the discriminant is critical in determining the nature of the roots without actually solving for them entirely.

Roots of Quadratic Equation

The roots of a quadratic equation are the x-values where the parabola (graph of the quadratic equation) intersects the x-axis. Graphically, these are where the function \(f(x) = ax^2 + bx + c\) gets the value zero. Calculating the roots is fundamental in algebra as it provides the solutions to the equation. In relation to our provided example, once the quadratic equation \(y^2 - 5y + 6 = 0\) is solved using the quadratic formula or by factoring, you get two roots, \(y_1 = 3\) and \(y_2 = 2\), which correspond to the points on the graph where the parabola intersects the x-axis.

It is essential to substitute back for y if you have initially substituted to simplify the equation, as in the exercise \(y=(x-2)\), to find the actual roots of the original quadratic equation. Therefore, \(x_1=5\) and \(x_2=4\) are the roots for the given equation \(\left(x-2\right)^2 - 5\left(x-2\right) + 6=0\). Performing a check by substituting these values back into the original equation verifies their accuracy.