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Chapter 1: Problem 52

Find the real solutions of each equation. $$ (2-x)^{2}+(2-x)-20=0 $$

Short Answer

Expert verified
The solutions are x = 7 and x = -2.

Step by step solution

01

- Substitute y

Let us denote the expression (2-x) as y . This substitution simplifies our equation to a quadratic form: y^2 + y - 20 = 0 .
02

- Factor the quadratic equation

We need to factor the quadratic equation y^2 + y - 20 = 0 . We look for two numbers that multiply to -20 and add up to 1 . These numbers are 5 and -4 . Hence, we can rewrite the equation as (y + 5)(y - 4) = 0 .
03

- Solve for y

Set each factor equal to zero to find the values of y : y + 5 = 0 gives y = -5 , and y - 4 = 0 gives y = 4 .
04

- Substitute back x

Recall that y = 2 - x . Substitute y back into the equations y = -5 and y = 4 to solve for x: 2 - x = -5 gives x = 7 , and 2 - x = 4 gives x = -2 .
05

- Verify the solutions

Substitute x = 7 and x = -2 back into the original equation to check if they satisfy it: For x = 7 , (2 - 7)^2 + (2 - 7) - 20 = 25 - 5 - 20 = 0 , which is true, and for x = -2 , (2 - (-2))^2 + (2 - (-2)) - 20 = 16 + 4 - 20 = 0 , which is also true.

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Key Concepts

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

Factoring Quadratics
Quadratic equations often appear in the form ax^2 + bx + c = 0 . To solve them by factoring, we need to express the quadratic equation as a product of two binomials set to zero. For our example, after the substitution method, we got the equation y^2 + y - 20 = 0 . Here is how we factor it step-by-step:
  • Identify two numbers that multiply to the constant term (-20) and add up to the coefficient of the linear term (1). Those numbers are 5 and -4.
  • Rewrite the quadratic equation using these two numbers: (y + 5)(y - 4) = 0 .
  • Set each factor equal to zero: y + 5 = 0 and y - 4 = 0 .
  • Solve for y : y = -5 and y = 4 .
Factoring quadratics is a powerful method to solve equations and finding the right pair of numbers is the key step. This approach works efficiently when the quadratic can be factored into integer values.
Substitution Method
The substitution method is particularly useful when the quadratic equation is not initially in the standard form ax^2 + bx + c = 0 . In our example, we started with the equation (2 - x)^2 + (2 - x) - 20 = 0 . To simplify this, we used substitution as follows:
  • Let y = 2 - x . This transforms the equation: y^2 + y - 20 = 0 .
  • We then solved y^2 + y - 20 = 0 as a quadratic equation.
  • After solving, we obtained y = -5 and y = 4 , and substituted back y = 2 - x to find x .
  • This yielded: 2 - x = -5 which led to x = 7 and 2 - x = 4 which led to x = -2 .
The substitution method effectively reduces and simplifies the original equation into a more manageable form, making it easier to solve.
Verification of Solutions
Verification of solutions ensures that the solutions we find actually satisfy the original equation. Here鈥檚 how to verify:
  • Substitute each solution back into the original equation: For x = 7 : Replace x with 7 in the equation (2 - 7)^2 + (2 - 7) - 20 = 0 .
    Calculating, we get (-5)^2 - 5 - 20 = 0 , which simplifies to 25 - 5 - 20 = 0 , confirming x = 7 as a valid solution.
  • Similarly for x = -2 : Replace x with -2 in the equation: (2 - (-2))^2 + (2 - (-2)) - 20 = 0 .
    Calculating, we get (4)^2 + 4 - 20 = 0 which simplifies to 16 + 4 - 20 = 0 , confirming x = -2 as a valid solution.
This process verifies that both x = 7 and x = -2 satisfy the original quadratic equation. Verification is crucial, as it confirms the validity of the solutions obtained.

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