Problem 89 - Algebra 2

Question

Solve using the Quadratic Formula. \(x^{2}-9 x+15=0\)

Step-by-Step Solution

Answer

The roots of the equation \(x^{2}-9 x+15=0\) are \(x = \frac{9 + \sqrt{21}}{2}\) and \(x = \frac{9 - \sqrt{21}}{2}\).

Step 1: Identify the coefficients

From the given quadratic equation \(x^{2}-9 x+15=0\), you can see that 'a' is 1 (the coefficient of \(x^{2}\)), 'b' is -9 (the coefficient of 'x'), and 'c' is 15.

Step 2: Substitute into the Quadratic Formula

Substitute 'a', 'b', and 'c' into the Quadratic Formula: \(x = \frac{-(-9) \pm \sqrt{(-9)^{2} - 4*1*15}}{2*1}\). This simplifies to: \(x = \frac{9 \pm \sqrt{81 - 60}}{2}\).

Step 3: Simplify the expression

Continue to simplify the equation: \(x = \frac{9 \pm \sqrt{21}}{2}\). So, the roots of the given quadratic equation are \(x = \frac{9 + \sqrt{21}}{2}\) and \(x = \frac{9 - \sqrt{21}}{2}\).

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree. It has the standard form:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants, and \( x \) represents an unknown variable. The coefficient \( a \) cannot be zero because, if it were, the equation would be linear, not quadratic.
Quadratic equations are widely used in algebra because they model many real-world phenomena, such as the path of a projectile. They have interesting mathematical properties because they can provide two potential solutions (roots) for the unknown variable \( x \).
Solving a quadratic equation means finding the values of \( x \) that make the equation true. Various methods can solve them, including factoring, completing the square, and the quadratic formula. Each method has distinct advantages depending on the specific equation at hand.

Roots of Equations

The roots of an equation are the values that satisfy the equation exactly, making it equal to zero. For quadratic equations, the roots can be real or complex numbers. A quadratic equation can have up to two roots because it is a second degree polynomial.

• If the roots are real and different, the parabola representing the equation intersects the x-axis at two points.
• If they are real and identical, the parabola touches the x-axis at one point, and we say the equation has a repeated root.
• If the roots are complex, the parabola doesn't touch the x-axis at all, and we consider the roots as a pair of complex conjugates.

For an equation like \( x^2 - 9x + 15 = 0 \), the roots are found using the quadratic formula method. This helps in finding both roots efficiently no matter their nature, provided they follow the laws of algebra and arithmetic.

Algebraic Solutions

Algebraic solutions refer to the methods and steps taken to find the roots of equations using algebra. For quadratic equations, the algebraic solutions typically involve concepts such as simplifying expressions and applying formulas.
The quadratic formula is a central tool for algebraic solutions of quadratic equations:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Applying this formula includes the following:

• Identify coefficients \( a \), \( b \), and \( c \) from the equation.
• Calculate the discriminant \( b^2 - 4ac \) which helps determine the nature of the roots.
• Perform operations of addition, subtraction, multiplication, and division to simplify expressions.

In the given problem, substituting \( a = 1 \), \( b = -9 \), and \( c = 15 \) into the formula reveals the roots. As you simplify using algebraic steps, you discover the solutions are \( x = \frac{9 + \sqrt{21}}{2} \) and \( x = \frac{9 - \sqrt{21}}{2} \).
Through these algebraic solutions, we achieve both numerical answers and a deeper understanding of the equation's behavior.