Problem 4 - Algebra 2 and Trigonometry

Question

Solve and check each of the equations. \(x^{2}-7 x+10=0\)

Step-by-Step Solution

Answer

The solutions are \(x = 5\) and \(x = 2\).

Step 1: Identifying the Type of Equation

Recognize that the equation \(x^2 - 7x + 10 = 0\) is a quadratic equation, which has the general form \(ax^2 + bx + c = 0\). For this equation, \(a = 1\), \(b = -7\), and \(c = 10\).

Step 2: Factoring the Quadratic Equation

To solve the quadratic equation, we first need to factor it. We look for two numbers that multiply to \(c = 10\) and add to \(b = -7\). The numbers \(-5\) and \(-2\) satisfy these conditions because \(-5 \times -2 = 10\) and \(-5 + (-2) = -7\). Thus, the equation can be factored as \((x - 5)(x - 2) = 0\).

Step 3: Solving Each Factor

Once the equation is factored, we set each factor equal to zero to find the solutions for \(x\). So, solve \(x - 5 = 0\) and \(x - 2 = 0\). Solving these gives \(x = 5\) and \(x = 2\).

Step 4: Checking the Solutions

Substitute each solution back into the original equation to verify correctness. First, check \(x = 5\): \((5)^2 - 7(5) + 10 = 0\) simplifies to \(25 - 35 + 10 = 0\), which is true. Second, check \(x = 2\): \((2)^2 - 7(2) + 10 = 0\) simplifies to \(4 - 14 + 10 = 0\), which is also true. Both solutions satisfy the equation.

Key Concepts

Factoring Quadratics

Factoring quadratics is a key skill in solving quadratic equations. The aim is to express the equation in a product of two binomials. For the quadratic equation \(x^2 - 7x + 10 = 0\), we start by identifying the coefficients with \(a = 1\), \(b = -7\), and \(c = 10\).

To factor this equation, we need to find two numbers that multiply to the constant term, \(c = 10\), and add up to the linear coefficient, \(b = -7\). These two numbers are \(-5\) and \(-2\), since \(-5 \times -2 = 10\) and \(-5 + (-2) = -7\).

Thus, the factored form of the equation is \((x - 5)(x - 2) = 0\). This step changes the problem from a quadratic into a simpler, linear form, making it easier to solve.

Solving Quadratic Equations

After factoring the quadratic equation, solving it becomes straightforward. We take each factor from the expression \((x - 5)(x - 2) = 0\) and set them equal to zero, one at a time. This is based on the zero product property, which states that if a product of two numbers is zero, then at least one of the numbers must be zero.

• First, solve for \(x - 5 = 0\). Add 5 to both sides to obtain \(x = 5\).
• Next, solve for \(x - 2 = 0\). Add 2 to both sides to obtain \(x = 2\).

These calculations provide the solutions to the original quadratic equation, meaning \(x = 5\) and \(x = 2\) are the points where the quadratic equation would cross the x-axis when graphed.

Checking Solutions of Equations

Once we've found potential solutions by factoring and solving, the next step is to verify them by substituting back into the original equation. This ensures our solutions are correct.

• For \(x = 5\): Substitute 5 into the equation: \((5)^2 - 7(5) + 10 = 0\). Simplify to obtain 25 - 35 + 10 = 0, which equals 0.
• For \(x = 2\): Substitute 2 into the equation: \((2)^2 - 7(2) + 10 = 0\). Simplify to obtain 4 - 14 + 10 = 0, also equaling 0.

Both checks return zero, verifying that \(x = 5\) and \(x = 2\) are indeed the correct solutions to the original quadratic equation. Ensuring solutions satisfy the original equation is an important step in validating our work.