Math Problem Statement

Solve the equation by using completing square method: 3(x^2 + 1) = 4(x + 1)

Solution

The equation provided is:

3(x2+1)=4(x+1)3(x^2 + 1) = 4(x + 1)

We are asked to solve this equation using the completing the square method. Let's go through the steps:

Step 1: Expand both sides

First, expand both sides of the equation:

3(x2+1)=4(x+1)3(x^2 + 1) = 4(x + 1)

Expanding both sides:

3x2+3=4x+43x^2 + 3 = 4x + 4

Step 2: Rearrange the terms

Move all terms to one side to set the equation to zero:

3x2+34x4=03x^2 + 3 - 4x - 4 = 0

Simplify:

3x24x1=03x^2 - 4x - 1 = 0

Step 3: Divide by 3 (if necessary)

If we want the coefficient of x2x^2 to be 1 for easier completing the square, divide the entire equation by 3:

x243x13=0x^2 - \frac{4}{3}x - \frac{1}{3} = 0

Step 4: Complete the square

To complete the square, we need to adjust the quadratic and linear terms. Take the coefficient of xx, which is 43-\frac{4}{3}, divide it by 2, and square it:

(432)2=(23)2=49\left(\frac{-\frac{4}{3}}{2}\right)^2 = \left(\frac{-2}{3}\right)^2 = \frac{4}{9}

Now add and subtract 49\frac{4}{9} to the equation:

x243x+494913=0x^2 - \frac{4}{3}x + \frac{4}{9} - \frac{4}{9} - \frac{1}{3} = 0

This can be written as:

(x23)24913=0\left(x - \frac{2}{3}\right)^2 - \frac{4}{9} - \frac{1}{3} = 0

Step 5: Simplify

Now combine the constant terms:

4913=4939=79-\frac{4}{9} - \frac{1}{3} = -\frac{4}{9} - \frac{3}{9} = -\frac{7}{9}

So the equation becomes:

(x23)2=79\left(x - \frac{2}{3}\right)^2 = \frac{7}{9}

Step 6: Solve for xx

Take the square root of both sides:

x23=±79=±73x - \frac{2}{3} = \pm \sqrt{\frac{7}{9}} = \pm \frac{\sqrt{7}}{3}

Now solve for xx:

x=23±73x = \frac{2}{3} \pm \frac{\sqrt{7}}{3}

Thus, the two solutions are:

x=2+73orx=273x = \frac{2 + \sqrt{7}}{3} \quad \text{or} \quad x = \frac{2 - \sqrt{7}}{3}

These are the solutions to the equation.


Would you like a more detailed explanation on any step?

Here are 5 related questions for further practice:

  1. How do you complete the square for an equation with a different constant on the right-hand side?
  2. Can the completing the square method be applied to equations with higher degrees?
  3. How does completing the square help derive the quadratic formula?
  4. What are the applications of completing the square in real-world problems?
  5. How do you solve a quadratic equation using the quadratic formula?

Tip: Completing the square is not just a method for solving quadratic equations, it also helps in rewriting equations for conic sections (like circles and parabolas).

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Completing the Square

Formulas

Quadratic equation formula: ax^2 + bx + c = 0
Completing the square: (x - h)^2 = k

Theorems

Quadratic formula

Suitable Grade Level

Grades 9-11