Problem 58 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \((x-4)^{2}+(x-4)-20=0\)

Step-by-Step Solution

Answer

The solutions are \(x = -1\) and \(x = 8\).

Step 1: Simplify the Equation

First, let’s define a new variable to simplify the equation. Let’s set \(y = x - 4\), so the equation becomes \(y^2 + y - 20 = 0\).

Step 2: Factor the Quadratic Equation

Next, factor the quadratic equation. We are looking for two numbers that add to 1 (the coefficient of y) and multiply to -20. These numbers are 5 and -4. So, \(y^2 + y - 20 = (y + 5)(y - 4) = 0\).

Step 3: Solve for y

Set each factor equal to zero to find the value of y: \(y + 5 = 0\) or \(y - 4 = 0\).So, \(y = -5\) or \(y = 4\).

Step 4: Substitute Back in for x

Recall that \(y = x - 4\). Substitute \(y = -5\) back in: \(-5 = x - 4\), which simplifies to \(x = -1\).Now, substitute \(y = 4\) back in: \(4 = x - 4\), which simplifies to \(x = 8\).

Step 5: Check the Solutions

Finally, substitute \(x = -1\) and \(x = 8\) back into the original equation to verify.For \(x = -1\): i. \((x - 4)^2 = (-1 - 4)^2 = 25\)ii. \((x - 4) = -5\)iii. \((x-4)^2 + (x-4) - 20 = 25 - 5 - 20 = 0\) (true)For \(x = 8\): i. \((x - 4)^2 = (8 - 4)^2 = 16\)ii. \((x - 4) = 4\)iii. \((x-4)^2 + (x-4) - 20 = 16 + 4 - 20 = 0\) (true)Both values satisfy the original equation.

Key Concepts

factoring quadratics

When it comes to solving quadratic equations, factoring is a crucial method to learn. Quadratic equations are in the form of latex: ax^2 + bx + c = 0d. To solve them, we look for two numbers that multiply to the constant term (c) and add up to the coefficient of the linear term (b).

• Firstly, we need to rewrite the equation in its standard form, if it’s not already.
• Identify the coefficients: a, b, and c.
• Find two numbers that multiply to 'ac' and add up to 'b'.
• Once identified, we can rewrite the quadratic as a product of two binomials.

In our exercise, we had the equation latex: y^2 + y - 20 = 0d.

• The coefficient 'a' is 1, 'b' is 1, and 'c' is -20.
• We need numbers that multiply to -20 and add to 1. We find 5 and -4 work.
• So, we factor it as latex: (y + 5)(y - 4) = 0d.

By solving each binomial factor set to zero, latex: y + 5 = 0d and latex: y - 4 = 0d, we get the solutions for 'y' which can be substituted back to find 'x'.

variable substitution

Variable substitution is a handy technique when dealing with complex equations. It allows simplification by introducing a new variable temporarily.
In our exercise, the original equation latex: (x-4)^2 + (x-4) - 20 = 0dcan be simplified by substituting latex: y = x - 4d.

• This transforms the complicated expression into an easier form: latex: y^2 + y - 20 = 0d.
• We then solve for 'y' in this simplified equation.
• Once we find the values for 'y', we reverse the substitution to get back to 'x'.

Variable substitution often turns a messy problem into a straightforward one, making it easier to solve and understand.

checking solutions

Always check your solutions to ensure they satisfy the original equation. This step verifies your work and ensures no mistakes were made.
Here’s how you can do it:

• Substitute each solution back into the original equation, not the simplified one.
• Calculate each term to see if the left-hand side equals the right-hand side of the equation.
• If both sides equal, your solution is correct. If not, recheck your steps.

In our example, we found solutions latex: x = -1d and latex: x = 8d by reversing the substitution.

• We substitute latex: x = -1d back into the original equation and verify each term sums to zero.
• Do the same with latex: x = 8d. Both solutions hold true, confirming they are correct.

Checking solutions prevents errors and boosts confidence in your problem-solving process.