Problem 84 - College Algebra

Question

Solve each equation in Exercises 73-98 by the method of your choice. \((2 x+7)^{2}=25\)

Step-by-Step Solution

Answer

Solving the equation \((2x+7)^{2}=25\), yields two possible solutions for \(x\), which are \(x= -1\) and \(x = -6\).

Step 1: Expand the squared term

In this phase, the equation is expanded. \((2x+7)^{2}=25\) becomes \(4x^{2} + 28x + 49=25\).

Step 2: Rearrange the equation

Now, rearrange the equation to one side by subtracting 25 from all sides to make the equation equal to zero. This leads to \(4x^{2} + 28x + 49 - 25 = 0\) and simplifies to \(4x^{2}+28x+24=0\).

Step 3: Factoring the equation

Factor the equation obtained in the last step. Take out the common factors, in this case, 4. So, \(4x^{2}+28x+24=0\) becomes \(4(x^{2}+7x+6)=0\). Which factors to \((x+1)(x+6)=0\).

Step 4: Solve for x

Having factored the equation out to \(4(x+1)(x+6)=0\), we can now solve for x by setting each factor equal to zero. x+1=0 gives x=-1 and x+6=0 yields x=-6.

Key Concepts

Factoring Quadratics

Factoring quadratics is an essential technique in algebra which involves breaking down polynomial expressions into products of simpler factors. In solving quadratic equations, factoring is often a preferred method when the polynomial can be decomposed easily. Consider the quadratic equation from the exercise \(4x^2+28x+24=0\). To factor this, we first look for common factors in each term.

In this case, each coefficient is divisible by 4. Factoring out this common term gives us \((x^2+7x+6)=0\). We then need to find two numbers that multiply to give the constant term (here, 6) and add up to give the coefficient of the \((x\) term, which is 7. For the quadratic \((x^2+7x+6\), these numbers are 1 and 6, thus factoring further into \((x+1)(x+6)=0\).

Quadratic Formula

The quadratic formula provides a straightforward way to solve any quadratic equation of the form \(ax^2+bx+c=0\). The formula is \((x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). It is especially useful when factoring is difficult or impractical. To apply the formula to our exercise, we would identify \(a=4\), \(b=28\), and \(c=24\), and then plug these values into the formula. However, considering we have already factored the equation, we can see that the quadratic formula would have given us the roots \(x=-1\) and \(x=-6\), which are the same solutions obtained through factoring.

Expanding Binomials

Expanding binomials is a key skill when working with quadratic equations. It involves multiplying two binomial expressions to form a quadratic expression. The initial step in our exercise involved expanding the squared binomial \(2x+7)^{2}\). This uses the foil method (First, Outer, Inner, Last), which yields \(4x^2 + 28x + 49\) after simplification. Understanding binomial expansion is crucial, as it helps in solving equations by first representing them in their standard quadratic form before applying other algebraic problem-solving methods.

Algebra Problem Solving

Algebraic problem solving encompasses a variety of techniques to solve equations, such as factoring, applying the quadratic formula, and expanding binomials, as seen in the exercise. To master algebra problem solving, one should understand when and how to apply these methods effectively. In our example, recognizing that the quadratic could be factored easily after expansion and rearrangement is part of strategic problem-solving. For more complicated quadratics that cannot be factored, one might opt for the quadratic formula. Each method has its place, and developing a good sense for appropriate application is key to becoming proficient in algebra problem solving.