Square of sum: Domain of definition

Equations with Variables on One SideUsing short multiplication formulasUsing variablesEquations with denominatorsIdentify the greater valueUsing fractionsUsing quadrilateralsComplete the missing numbersMore than one factorizationSystem of equations with no solutionData with powers and rootsUsing multiple rulesNumber of termsSolving the problemEquations with variables on both sidesTransition between expressionsResulting in a quadratic equation
Question 1

Solve the following equation:

\( \frac{x^3+1}{(x+1)^2}=x \)

Question 2

Solve the following equation:

\( \frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x \)

Examples with solutions for Square of sum: Domain of definition

Exercise #1

Solve the following equation:

x3+1(x+1)2=x \frac{x^3+1}{(x+1)^2}=x

Video Solution

Step-by-Step Solution

To solve the equation x3+1(x+1)2=x \frac{x^3 + 1}{(x+1)^2} = x , we will follow these steps:

  • Step 1: Set up the equation for solving by cross-multiplying.
  • Step 2: Simplify and solve the resulting polynomial equation.
  • Step 3: Solve for the values of x x .

Let's work through the solution:

Step 1: Cross-multiply to eliminate the fraction:

(x3+1)=x⋅(x+1)2 (x^3 + 1) = x \cdot (x+1)^2

Expand the right-hand side:

x⋅(x2+2x+1)=x3+2x2+x x \cdot (x^2 + 2x + 1) = x^3 + 2x^2 + x

Step 2: Set the expanded equation equal:

x3+1=x3+2x2+x x^3 + 1 = x^3 + 2x^2 + x

Cancel x3 x^3 from both sides:

1=2x2+x 1 = 2x^2 + x

Re-arrange the equation to form a standard quadratic equation:

0=2x2+x−1 0 = 2x^2 + x - 1

Step 3: Solve the quadratic equation using the quadratic formula:

Here, a=2, b=1, c=−1.\text{Here, } a = 2, \, b = 1, \, c = -1.

The quadratic formula is:

x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the values of a a , b b , and c c into the formula:

x=−1±12−4×2×(−1)2×2 x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2}

Calculate the discriminant and simplify:

x=−1±1+84=−1±94 x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4}

Simplify further:

x=−1±34 x = \frac{-1 \pm 3}{4}

This gives the solutions:

x=24=12 x = \frac{2}{4} = \frac{1}{2} x=−44=−1 x = \frac{-4}{4} = -1

Since x=−1 x = -1 would make the denominator zero, it is not allowed as a solution. Thus, the only valid solution is:

Therefore, the solution to the equation is x=12 x = \frac{1}{2} .

Answer

x=12 x=\frac{1}{2}

Exercise #2

Solve the following equation:

(2x+1)2x+2+(x+2)22x+1=4.5x \frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x

Step-by-Step Solution

In order to solve the equation, start by removing the denominators.

To do this, we'll multiply the denominators:

(2x+1)2⋅(2x+1)+(x+2)2⋅(x+2)=4.5x(2x+1)(x+2) (2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)

Open the parentheses on the left side, making use of the distributive property:

(4x2+4x+1)⋅(2x+1)+(x2+4x+4)⋅(x+2)=4.5x(2x+1)(x+2) (4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)

Continue to open the parentheses on the right side of the equation:

(4x2+4x+1)⋅(2x+1)+(x2+4x+4)⋅(x+2)=4.5x(2x2+5x+2) (4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)

Simplify further:

(4x2+4x+1)⋅(2x+1)+(x2+4x+4)⋅(x+2)=9x3+22.5x+9x (4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x

Go back and simplify the parentheses on the left side of the equation:

8x3+8x2+2x+4x2+4x+1+x3+4x2+4x+2x2+8x+8=9x3+22.5x+9x 8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x

Combine like terms:

9x3+18x2+18x+9=9x3+22.5x+9x 9x^3+18x^2+18x+9=9x^3+22.5x+9x

Notice that all terms can be divided by 9 as shown below:

x3+2x2+2x+1=x3+2.5x+x x^3+2x^2+2x+1=x^3+2.5x+x

Move all numbers to one side:

x3−x3+2x2−2.5x2+2x−x+9=0 x^3-x^3+2x^2-2.5x^2+2x-x+9=0

We obtain the following:

0.5x2−x−1=0 0.5x^2-x-1=0

In order to remove the one-half coefficient, multiply the entire equation by 2

x2−2x−2=0 x^2-2x-2=0

Apply the square root formula, as shown below-

2±122 \frac{2±\sqrt{12}}{2}

Apply the properties of square roots in order to simplify the square root of 12:

2±232 \frac{2±2\sqrt{3}}{2} Divide both the numerator and denominator by 2 as follows:

1±3 1±\sqrt{3}

Answer

x=1±3 x=1±\sqrt{3}