Solving Equations Using Root Formulas A Comprehensive Guide

Hey guys! Let's dive into the fascinating world of solving equations using root formulas. This guide will walk you through several examples, breaking down each step to make it super easy to understand. We'll tackle everything from rational equations to quadratic equations, ensuring you're well-equipped to handle any algebraic challenge that comes your way. So, grab your pencils and let's get started!

Understanding Root Formulas

Before we jump into the problems, it's important to understand what root formulas are and why they're so useful. In mathematics, a "root" of an equation is a value that, when plugged into the equation, makes the equation true. Root formulas provide a systematic way to find these values, especially for polynomial equations. For quadratic equations (those of the form ax² + bx + c = 0), the quadratic formula is our go-to tool. For higher-degree polynomials, finding roots can be more complex, often involving factoring, synthetic division, or numerical methods. The beauty of root formulas lies in their ability to provide exact solutions, which is crucial in many mathematical and scientific applications. The quadratic formula, for instance, elegantly expresses the roots in terms of the coefficients a, b, and c, giving us a direct path to the solutions without the guesswork of trial and error. This is particularly valuable when dealing with equations that don't factor easily or have irrational roots. Understanding the underlying principles of root formulas not only helps in solving equations but also deepens our understanding of the structure and behavior of polynomial functions. So, let's keep this in mind as we move forward and apply these concepts to solve some exciting problems!

(a) Solving a Rational Equation

Let's kick things off with a rational equation: $\frac{3 x-7}{2 x-5}=\frac{x+1}{x-1}$.

Step 1: Cross-Multiplication

To get rid of the fractions, we'll cross-multiply. This means multiplying the numerator of the first fraction by the denominator of the second and vice versa. So, we get:

$$(3x - 7)(x - 1) = (x + 1)(2x - 5)$$

Step 2: Expanding the Products

Next, we need to expand both sides of the equation. Let's use the distributive property (aka FOIL method) to do this:

$$3x^2 - 3x - 7x + 7 = 2x^2 - 5x + 2x - 5$$

Step 3: Simplifying the Equation

Now, let's simplify by combining like terms on both sides:

$$3x^2 - 10x + 7 = 2x^2 - 3x - 5$$

Step 4: Rearranging to a Quadratic Form

To solve this, we need to set the equation to zero. Subtract the right side from the left side:

$$3x^2 - 10x + 7 - (2x^2 - 3x - 5) = 0$$

$$3x^2 - 10x + 7 - 2x^2 + 3x + 5 = 0$$

$$x^2 - 7x + 12 = 0$$

Step 5: Factoring the Quadratic Equation

Now we have a quadratic equation in the form ax² + bx + c = 0. Let's try to factor it. We're looking for two numbers that multiply to 12 and add up to -7. Those numbers are -3 and -4.

So, we can factor the equation as:

$$(x - 3)(x - 4) = 0$$

Step 6: Finding the Roots

To find the roots, set each factor equal to zero:

$$x - 3 = 0 \text{ or } x - 4 = 0$$

Solving for x, we get:

$$x = 3 \text{ or } x = 4$$

Step 7: Checking for Extraneous Solutions

It's crucial to check our solutions in the original equation, especially with rational equations, to make sure we don't have any extraneous solutions (solutions that don't actually work). Let's plug x = 3 and x = 4 back into the original equation.

For x = 3:

$$\frac{3(3)-7}{2(3)-5} = \frac{3+1}{3-1}$$

$$\frac{9-7}{6-5} = \frac{4}{2}$$

$$\frac{2}{1} = 2$$

$$2 = 2$$

So, x = 3 is a valid solution.

For x = 4:

$$\frac{3(4)-7}{2(4)-5} = \frac{4+1}{4-1}$$

$$\frac{12-7}{8-5} = \frac{5}{3}$$

$$\frac{5}{3} = \frac{5}{3}$$

So, x = 4 is also a valid solution.

Final Answer

The solutions to the equation $\frac{3 x-7}{2 x-5}=\frac{x+1}{x-1}$ are x = 3 and x = 4.

(b) Solving Another Rational Equation

Now, let's tackle another rational equation: $\frac{20+x^2}{15}=11-\frac{3 x^2+5}{10}$.

Step 1: Eliminating the Fractions

To get rid of the fractions, we'll find the least common multiple (LCM) of the denominators, which are 15 and 10. The LCM of 15 and 10 is 30. We'll multiply both sides of the equation by 30:

$$30 \cdot \frac{20+x^2}{15} = 30 \cdot \left(11-\frac{3 x^2+5}{10}\right)$$

Step 2: Distributing and Simplifying

Now, distribute the 30 on both sides:

$$2(20 + x^2) = 30 \cdot 11 - 3(3x^2 + 5)$$

Expand the terms:

$$40 + 2x^2 = 330 - 9x^2 - 15$$

Step 3: Rearranging the Equation

Let's simplify and rearrange the equation to get all terms on one side:

$$40 + 2x^2 = 315 - 9x^2$$

Move all terms to the left side:

$$2x^2 + 9x^2 + 40 - 315 = 0$$

$$11x^2 - 275 = 0$$

Step 4: Isolating x²

Now, let's isolate x²:

$$11x^2 = 275$$

Divide both sides by 11:

$$x^2 = 25$$

Step 5: Finding the Roots

To find x, take the square root of both sides. Remember to consider both positive and negative roots:

$$x = \pm\sqrt{25}$$

$$x = \pm 5$$

Final Answer

The solutions to the equation $\frac{20+x^2}{15}=11-\frac{3 x^2+5}{10}$ are x = 5 and x = -5.

(c) Solving a Quadratic Equation with Parameters

Let's move on to a quadratic equation with parameters: $x^2(a-b)-x(a+b)+2 b=0$.

Step 1: Identifying Coefficients

First, let's identify the coefficients a, b, and c in the quadratic equation Ax² + Bx + C = 0:

• A = (a - b)
• B = -(a + b)
• C = 2b

Step 2: Using the Quadratic Formula

Now, we'll use the quadratic formula to find the roots:

$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$

Plug in the coefficients:

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

Step 3: Simplifying the Expression

Simplify the expression step by step:

$$x = \frac{(a + b) \pm \sqrt{(a+b)^2 - 8b(a-b)}}{2(a-b)}$$

Expand the terms under the square root:

$$x = \frac{(a + b) \pm \sqrt{a^2 + 2ab + b^2 - 8ab + 8b^2}}{2(a-b)}$$

$$x = \frac{(a + b) \pm \sqrt{a^2 - 6ab + 9b^2}}{2(a-b)}$$

Step 4: Recognizing a Perfect Square

Notice that the expression under the square root is a perfect square:

$$a^2 - 6ab + 9b^2 = (a - 3b)^2$$

So, we can simplify further:

$$x = \frac{(a + b) \pm \sqrt{(a - 3b)^2}}{2(a-b)}$$

$$x = \frac{(a + b) \pm (a - 3b)}{2(a-b)}$$

Step 5: Finding the Two Roots

Now, let's find the two roots by considering the plus and minus signs:

For the plus sign:

$$x_1 = \frac{(a + b) + (a - 3b)}{2(a-b)}$$

$$x_1 = \frac{2a - 2b}{2(a-b)}$$

$$x_1 = \frac{2(a - b)}{2(a-b)}$$

$$x_1 = 1$$

For the minus sign:

$$x_2 = \frac{(a + b) - (a - 3b)}{2(a-b)}$$

$$x_2 = \frac{a + b - a + 3b}{2(a-b)}$$

$$x_2 = \frac{4b}{2(a-b)}$$

$$x_2 = \frac{2b}{a-b}$$

Final Answer

The solutions to the equation $x^2(a-b)-x(a+b)+2 b=0$ are x = 1 and $x = \frac{2b}{a-b}$.

(d) Solving a Quadratic Equation with Fractions

Finally, let's solve the equation $x(x+\frac{b}{a})=-\frac{b^2}{4 a}$.

Step 1: Expanding and Rearranging

First, expand the left side and rearrange the equation to get it into the standard quadratic form:

$$x^2 + \frac{b}{a}x = -\frac{b^2}{4a}$$

Move the term on the right to the left side:

$$x^2 + \frac{b}{a}x + \frac{b^2}{4a} = 0$$

Step 2: Identifying Coefficients

Now, identify the coefficients A, B, and C:

• A = 1
• B = b/a
• C = b² / (4a)

Step 3: Using the Quadratic Formula

Apply the quadratic formula:

$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$

Plug in the coefficients:

$$x = \frac{-\frac{b}{a} \pm \sqrt{\left(\frac{b}{a}\right)^2 - 4(1)\left(\frac{b^2}{4a}\right)}}{2(1)}$$

Step 4: Simplifying the Expression

Simplify the expression under the square root:

$$x = \frac{-\frac{b}{a} \pm \sqrt{\frac{b^2}{a^2} - \frac{b^2}{a}}}{2}$$

To combine the terms under the square root, we need a common denominator, which is a²:

$$x = \frac{-\frac{b}{a} \pm \sqrt{\frac{b^2}{a^2} - \frac{ab^2}{a^2}}}{2}$$

Something seems off here. Let's recheck our steps. Ah, there's a mistake! It should be:

$$x = \frac{-\frac{b}{a} \pm \sqrt{\frac{b^2}{a^2} - \frac{4b^2}{4a}}}{2}$$

$$x = \frac{-\frac{b}{a} \pm \sqrt{\frac{b^2}{a^2} - \frac{b^2}{a}}}{2}$$

We made a mistake in simplifying. Let’s correct it. The term 4AC should be: 4 * 1 * (b^2 / 4a) = b^2 / a. So, under the square root, we have:

$$\frac{b^2}{a^2} - \frac{b^2}{a} = \frac{b^2 - ab^2}{a^2}$$

This still doesn't look right. Let’s go back to the original equation and try completing the square instead.

Step 1: Original Equation

$$x^2 + \frac{b}{a}x = -\frac{b^2}{4a}$$

Step 2: Completing the Square

To complete the square, we need to add $\left(\frac{b}{2a}\right)^2$ to both sides:

$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{b^2}{4a} + \left(\frac{b}{2a}\right)^2$$

$$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{b^2}{4a} + \frac{b^2}{4a^2}$$

Step 3: Simplifying

Now, the left side is a perfect square:

$$\left(x + \frac{b}{2a}\right)^2 = -\frac{b^2}{4a} + \frac{b^2}{4a^2}$$

Get a common denominator on the right side (4a²):

$$\left(x + \frac{b}{2a}\right)^2 = \frac{-ab^2 + b^2}{4a^2}$$

Step 4: Further Simplification

We have another mistake! The correct common denominator step should be:

$$\left(x + \frac{b}{2a}\right)^2 = \frac{-ab^2 + b^2}{4a^2}$$

It seems we are making this more complicated than it needs to be. Let’s go back to the step where we had:

$$x^2 + \frac{b}{a}x + \frac{b^2}{4a} = 0$$

Multiply the entire equation by 4a to clear the fraction (this is the easiest way!):

$$4ax^2 + 4bx + b^2 = 0$$

Step 5: Factoring

Now, we can see that this is a perfect square trinomial:

$$(2ax + b)^2 = 0$$

Step 6: Solving for x

Take the square root of both sides:

$$2ax + b = 0$$

Solve for x:

$$2ax = -b$$

$$x = -\frac{b}{2a}$$

Final Answer

The solution to the equation $x(x+\frac{b}{a})=-\frac{b^2}{4 a}$ is $x = -\frac{b}{2a}$.

Conclusion

Alright, guys, we've walked through solving several different types of equations using root formulas. From rational equations to quadratics with parameters, we've covered a lot of ground. Remember, the key is to break down each problem into manageable steps, double-check your work, and don't be afraid to try a different approach if you get stuck. Keep practicing, and you'll become a pro at solving equations in no time! Happy solving!