Problem 60 - Elementary Algebra

Question

Solve. $$ -2 x(x-9)=x+21 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{3}{2}\) and \(x = 7\).

Step 1: Expand the Left Side

Begin by expanding the left side of the equation \(-2x(x-9)\). This involves distributing \(-2x\) across the terms inside the parentheses:\[-2x(x-9) = -2x^2 + 18x\].So, the equation becomes:\[-2x^2 + 18x = x + 21\].

Step 2: Move All Terms to One Side

To simplify solving, move all terms to one side of the equation:\[-2x^2 + 18x - x - 21 = 0\].Combine like terms:\[-2x^2 + 17x - 21 = 0\].

Step 3: Solve the Quadratic Equation

Now, solve the quadratic equation \(-2x^2 + 17x - 21 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \( a = -2\), \( b = 17\), and \( c = -21\):First, calculate the discriminant \(b^2 - 4ac = 17^2 - 4(-2)(-21)\):\[= 289 - 168 = 121\].Use the quadratic formula:\[x = \frac{-17 \pm \sqrt{121}}{-4}\].Simplify further:\[x = \frac{-17 \pm 11}{-4}\].

Step 4: Calculate the Roots

Find the two possible values of \(x\).First root:\[x = \frac{-17 + 11}{-4} = \frac{-6}{-4} = \frac{3}{2}\].Second root:\[x = \frac{-17 - 11}{-4} = \frac{-28}{-4} = 7\].The solutions are \(x = \frac{3}{2}\) and \(x = 7\).

Key Concepts

Expansion of Expressions

When dealing with an expression, especially one involving parentheses, expansion is a crucial step. It means multiplying each term inside parentheses by the factor outside. In our exercise, we had to expand \(-2x(x-9)\). This is done by multiplying \(-2x\) by both \('x'\) and \('-9'\). Consequently, it becomes \(-2x^2 + 18x\).
Understanding this expansion helps in transforming complex expressions into simpler ones that are easier to manage. The distributive property is key here:

• Multiply the term outside the parentheses with each term inside.
• Simplify by combining like terms if possible.

This step sets the foundation for moving terms and eventually solving the equation by getting \-2x^2 + 18x = x + 21\.

Discriminant

The discriminant is part of the quadratic formula that tells us the nature of the roots of a quadratic equation. It's represented by the expression \('b^2 - 4ac'\), where a, b, and c are coefficients from the equation \('ax^2 + bx + c = 0'\).
In this context:

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there is exactly one real root.
• If it's negative, no real roots exist (but two complex roots do).

For our exercise, the calculation yielded \('121'\), a positive number, indicating two real roots. Thus, finding the discriminant helps us understand how many solutions to expect and whether they are real or complex.

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations, given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows for finding solutions of any quadratic equation \('ax^2 + bx + c = 0'\) directly. In our problem, the coefficients were \(a = -2, b = 17,\) and \(c = -21\).
After calculating the discriminant, the formula provided the roots:

• Plug into the formula to get two potential solutions: \(-b + \sqrt{b^2 - 4ac}\) and \(-b - \sqrt{b^2 - 4ac}\).
• Divide by \(2a\) to get the exact roots.

Using the quadratic formula ensures accurate solutions, even when factoring isn't apparent or possible.

Solving Equations

Solving quadratic equations involves finding the values of the variable that make the equation true. Each step from expansion to using the quadratic formula works towards this goal. In this exercise:

• We started with expanding to eliminate parentheses, simplifying the expression.
• Then, we moved terms around to form a standard quadratic equation.
• Calculating the discriminant helped predict the number and type of solutions.
• Finally, the quadratic formula determined the precise solutions.

Once we applied all these steps, we found two solutions for the variable:

• \(x = \frac{3}{2}\)
• \(x = 7\)

Thus, solving equations, especially quadratics, involves a series of systematic steps to transition from problem to solution effectively.