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Solve each equation. $$(3 w+2)^{2}-(w-5)^{2}=0$$

Short Answer

Expert verified
The solutions for the equation \((3w+2)^2 - (w-5)^2 = 0\) are \(w_1 \approx -0.77\) and \(w_2 \approx -3.48\).

Step by step solution

01

Expand the squares

First, we have to expand both squares. Remember that \((a \pm b)^2 = a^2 \pm 2ab + b^2\). Expanding the squares, we get: \[(3w+2)^2 - (w-5)^2 = (3w)^2 + 2\cdot3w\cdot2 + 2^2 - (w^2 - 2\cdot5\cdot w + 5^2)\]
02

Simplify the equation

Now, we need to simplify the equation by performing the arithmetic operations: \begin{align*} (3w)^2 + 2\cdot3w\cdot2 + 2^2 - (w^2 - 2\cdot5\cdot w + 5^2) &= 9w^2 + 12w + 4 - w^2 + 10w - 25 \\ &= 8w^2 + 22w - 21 \end{align*} Now our equation looks like this: \(8w^2 + 22w - 21 = 0\)
03

Factor the quadratic

Try to factor the quadratic equation: Unfortunately, there isn't an easy way to factor this quadratic. So, we move on to the quadratic formula.
04

Use the quadratic formula

Since the quadratic equation cannot be factored, we will use the quadratic formula to solve for \(w\). The quadratic formula is given by: \[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In our case, we have the equation \(8w^2 + 22w - 21 = 0\), so \(a = 8\), \(b = 22\), and \(c = -21\). Plugging in these values, we get: \begin{align*} w &= \frac{-22 \pm \sqrt{22^2 - 4\cdot8\cdot(-21)}}{2\cdot8} \\ &= \frac{-22 \pm \sqrt{484+672}}{16} \\ &= \frac{-22 \pm \sqrt{1156}}{16} \end{align*}
05

Calculate the solutions for \(w\)

Compute the final solutions for \(w\): \begin{align*} w_1 &= \frac{-22 + \sqrt{1156}}{16} \approx -0.77 \\ w_2 &= \frac{-22 - \sqrt{1156}}{16} \approx -3.48 \end{align*} So, the solutions for the equation are \(w_1 \approx -0.77\) and \(w_2 \approx -3.48\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
The Quadratic Formula is an essential tool for solving quadratic equations when factoring isn't possible. It provides a way to find the roots of any quadratic equation in the form \(ax^2 + bx + c = 0\). The formula is:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
The components \(a\), \(b\), and \(c\) represent the coefficients of the equation, with \(a\) being the coefficient of \(x^2\), \(b\) the coefficient of \(x\), and \(c\) the constant term.
Using this formula involves substituting these coefficients into the equation and using basic arithmetic to simplify. This is particularly useful when the quadratic equation doesn't factor easily.
In our problem, the equation \(8w^2 + 22w - 21 = 0\) couldn't be factored, so we applied the quadratic formula to find the values of \(w\) that satisfy the equation.
After calculating, we found approximate solutions for \(w\) as \(w_1 \approx -0.77\) and \(w_2 \approx -3.48\). This formula is especially useful because it works universally for all real number solutions.
Factoring Quadratics
Factoring is a common method for solving quadratics but only works when the quadratic equation can be easily decomposed into two linear expressions. For a quadratic in the form \(ax^2 + bx + c\), the goal is to find two numbers that multiply to \(ac\) and add to \(b\).
  • For example, the quadratic \(x^2 + 5x + 6 = 0\) can be factored into \((x + 2)(x + 3) = 0\).
Once factored, the solutions for the quadratic are the roots of each factor set to zero.
In our exercise, we attempted to factor the equation \(8w^2 + 22w - 21 = 0\). However, finding such numbers for this equation was not straightforward, leading us to use the quadratic formula instead.
Always consider attempting to factor before moving on to other methods, as it can provide an easy and quick solution path in suitable cases.
Solving Algebraic Equations
Solving algebraic equations involves finding the value(s) of the variable that make the equation true. This requires simplifying and manipulating the equation until the variable is isolated.
In the problem given, we started with the equation \((3w+2)^2-(w-5)^2=0\). The first step was expanding the squares to simplify the expression.
This process led to a simpler quadratic equation \(8w^2 + 22w - 21 = 0\), which we could tackle using the quadratic formula to find the solutions. It showcased the need for understanding multiple strategies like factoring and using formulas to address equations effectively.
  • Initially simplify the expressions by expanding or combining like terms
  • Use inverse operations to isolate and solve for variables
  • Apply appropriate mathematical tools, like the Quadratic Formula, when other methods do not yield results
Each step builds upon the previous, highlighting the sequential nature of solving these algebraic puzzles.
Polynomial Expansion
Polynomial expansion is a technique used to convert a product of binomials into an expanded polynomial form. This often involves using special formulas or identities.
For binomials, the expansion process relies on formulas like:
  • \((a + b)^2 = a^2 + 2ab + b^2\)
  • \((a - b)^2 = a^2 - 2ab + b^2\)
During the expansion process, each term in the first polynomial is multiplied by each term in the second polynomial. This method was applied to expand the given expression in the exercise before simplifying it into a quadratic form.
It's a critical skill because it underpins the simplification of complex algebraic expressions and leads to a form that can be solved using methods like factoring or applying the quadratic formula.
Understanding how to expand polynomials properly gains from knowing common polynomial identities, which aid in quickly rewriting expressions in a usable form for further solving steps.

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