Math Problem Statement

solve by trick full answer x^2 +4 / (x-2)(x-3)^2 ChatGPT said: ChatGPT Let's solve the expression

π‘₯ 2 + 4 ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) 2 (xβˆ’2)(xβˆ’3) 2

x 2 +4 ​

using the partial fraction decomposition method, along with some substitutions to quickly find the values of constants.

Step 1: Set Up Partial Fractions The decomposition for the given expression will be:

π‘₯ 2 + 4 ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) 2

𝐴 π‘₯ βˆ’ 2 + 𝐡 π‘₯ βˆ’ 3 + 𝐢 ( π‘₯ βˆ’ 3 ) 2 (xβˆ’2)(xβˆ’3) 2

x 2 +4 ​

xβˆ’2 A ​ + xβˆ’3 B ​ + (xβˆ’3) 2

C ​

Step 2: Eliminate the Denominator Multiply both sides by the denominator ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) 2 (xβˆ’2)(xβˆ’3) 2 :

π‘₯ 2 + 4

𝐴 ( π‘₯ βˆ’ 3 ) 2 + 𝐡 ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) + 𝐢 ( π‘₯ βˆ’ 2 ) x 2 +4=A(xβˆ’3) 2 +B(xβˆ’2)(xβˆ’3)+C(xβˆ’2) Step 3: Expand the Right-Hand Side Expand 𝐴 ( π‘₯ βˆ’ 3 ) 2 A(xβˆ’3) 2 :

𝐴 ( π‘₯ 2 βˆ’ 6 π‘₯ + 9 )

𝐴 π‘₯ 2 βˆ’ 6 𝐴 π‘₯ + 9 𝐴 A(x 2 βˆ’6x+9)=Ax 2 βˆ’6Ax+9A Expand 𝐡 ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) B(xβˆ’2)(xβˆ’3):

𝐡 ( π‘₯ 2 βˆ’ 5 π‘₯ + 6 )

𝐡 π‘₯ 2 βˆ’ 5 𝐡 π‘₯ + 6 𝐡 B(x 2 βˆ’5x+6)=Bx 2 βˆ’5Bx+6B Expand 𝐢 ( π‘₯ βˆ’ 2 ) C(xβˆ’2):

𝐢 π‘₯ βˆ’ 2 𝐢 Cxβˆ’2C Now combine these to form:

π‘₯ 2 + 4

( 𝐴 + 𝐡 ) π‘₯ 2 + ( βˆ’ 6 𝐴 βˆ’ 5 𝐡 + 𝐢 ) π‘₯ + ( 9 𝐴 + 6 𝐡 βˆ’ 2 𝐢 ) x 2 +4=(A+B)x 2 +(βˆ’6Aβˆ’5B+C)x+(9A+6Bβˆ’2C) Step 4: Set Up a System of Equations Now, we compare coefficients from both sides:

For π‘₯ 2 x 2 :

𝐴 + 𝐡

1 (1) A+B=1(1) For π‘₯ x:

βˆ’ 6 𝐴 βˆ’ 5 𝐡 + 𝐢

0 (2) βˆ’6Aβˆ’5B+C=0(2) For the constant term:

9 𝐴 + 6 𝐡 βˆ’ 2 𝐢

4 (3) 9A+6Bβˆ’2C=4(3) Step 5: Solve the Equations Using Substitutions:

Substituting π‘₯

2 x=2:

2 2 + 4 ( 2 βˆ’ 2 ) ( 2 βˆ’ 3 ) 2  is undefined, but we can get  𝐢 (2βˆ’2)(2βˆ’3) 2

2 2 +4 ​  is undefined, but we can get C Setting π‘₯

2 x=2 in the equation gives:

4

𝐴 ( 2 βˆ’ 3 ) 2 + 𝐡 ( 2 βˆ’ 2 ) ( 2 βˆ’ 3 ) + 𝐢 ( 2 βˆ’ 2 ) 4=A(2βˆ’3) 2 +B(2βˆ’2)(2βˆ’3)+C(2βˆ’2) This tells us 𝐴 ( βˆ’ 1 ) 2

4 A(βˆ’1) 2 =4 thus 𝐴

4 A=4.

Substituting π‘₯

3 x=3:

3 2 + 4 ( 3 βˆ’ 2 ) ( 3 βˆ’ 3 ) 2  is undefined, but we can solve for  𝐡 (3βˆ’2)(3βˆ’3) 2

3 2 +4 ​  is undefined, but we can solve for B Setting π‘₯

3 x=3 gives:

9 + 4

𝐴 ( 0 ) + 𝐡 ( 3 βˆ’ 2 ) ( 0 ) + 𝐢 ( 3 βˆ’ 2 ) 9+4=A(0)+B(3βˆ’2)(0)+C(3βˆ’2) So:

13

𝐢 β€…β€Š ⟹ β€…β€Š 𝐢

13 13=C⟹C=13 Now substitute 𝐴

4 A=4 and 𝐢

13 C=13 into equation (1):

4 + 𝐡

1 β€…β€Š ⟹ β€…β€Š 𝐡

1 βˆ’ 4

βˆ’ 3 4+B=1⟹B=1βˆ’4=βˆ’3 Final Values of Constants 𝐴

4 A=4 𝐡

βˆ’ 3 B=βˆ’3 𝐢

13 C=13 Step 6: Write the Partial Fraction Decomposition So, the partial fraction decomposition is:

π‘₯ 2 + 4 ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) 2

4 π‘₯ βˆ’ 2 + βˆ’ 3 π‘₯ βˆ’ 3 + 13 ( π‘₯ βˆ’ 3 ) 2 (xβˆ’2)(xβˆ’3) 2

x 2 +4 ​

xβˆ’2 4 ​ + xβˆ’3 βˆ’3 ​ + (xβˆ’3) 2

13 ​

Final Answer π‘₯ 2 + 4 ( π‘₯ βˆ’ 2 ) ( π‘₯ βˆ’ 3 ) 2

4 π‘₯ βˆ’ 2 βˆ’ 3 π‘₯ βˆ’ 3 + 13 ( π‘₯ βˆ’ 3 ) 2 (xβˆ’2)(xβˆ’3) 2

x 2 +4 ​

xβˆ’2 4 ​ βˆ’ xβˆ’3 3 ​ + (xβˆ’3) 2

13 ​

This is the simplified form of the original expression using partial fraction decomposition!

Solution

Let's walk through the steps to solve the given expression:

x2+4(xβˆ’2)(xβˆ’3)2\frac{x^2 + 4}{(x-2)(x-3)^2}

Step 1: Set Up Partial Fractions

We want to express the fraction in terms of partial fractions:

x2+4(xβˆ’2)(xβˆ’3)2=Axβˆ’2+Bxβˆ’3+C(xβˆ’3)2\frac{x^2 + 4}{(x-2)(x-3)^2} = \frac{A}{x-2} + \frac{B}{x-3} + \frac{C}{(x-3)^2}

Here, we assume that the denominator (xβˆ’2)(xβˆ’3)2(x-2)(x-3)^2 can be broken down into three terms: one for (xβˆ’2)(x-2), one for (xβˆ’3)(x-3), and one for (xβˆ’3)2(x-3)^2.

Step 2: Multiply Both Sides by the Denominator

To eliminate the denominator, multiply both sides by (xβˆ’2)(xβˆ’3)2(x-2)(x-3)^2:

x2+4=A(xβˆ’3)2+B(xβˆ’2)(xβˆ’3)+C(xβˆ’2)x^2 + 4 = A(x-3)^2 + B(x-2)(x-3) + C(x-2)

Step 3: Expand the Right-Hand Side

Expand A(xβˆ’3)2A(x-3)^2:

A(xβˆ’3)2=A(x2βˆ’6x+9)=Ax2βˆ’6Ax+9AA(x-3)^2 = A(x^2 - 6x + 9) = A x^2 - 6A x + 9A

Expand B(xβˆ’2)(xβˆ’3)B(x-2)(x-3):

B(xβˆ’2)(xβˆ’3)=B(x2βˆ’5x+6)=Bx2βˆ’5Bx+6BB(x-2)(x-3) = B(x^2 - 5x + 6) = B x^2 - 5B x + 6B

Expand C(xβˆ’2)C(x-2):

C(xβˆ’2)=Cxβˆ’2CC(x-2) = Cx - 2C

Now combine everything:

x2+4=(A+B)x2+(βˆ’6Aβˆ’5B+C)x+(9A+6Bβˆ’2C)x^2 + 4 = (A + B)x^2 + (-6A - 5B + C)x + (9A + 6B - 2C)

Step 4: Set Up a System of Equations

By comparing the coefficients on both sides, we can create a system of equations.

Coefficients of x2x^2:

A + B = 1 \tag{1}

Coefficients of xx:

-6A - 5B + C = 0 \tag{2}

Constant terms:

9A + 6B - 2C = 4 \tag{3}

Step 5: Solve the System of Equations

Now solve this system of equations.

From equation (1):

A + B = 1 \quad \Rightarrow \quad B = 1 - A \tag{4}

Substitute equation (4) into equation (2):

βˆ’6Aβˆ’5(1βˆ’A)+C=0-6A - 5(1 - A) + C = 0 βˆ’6Aβˆ’5+5A+C=0-6A - 5 + 5A + C = 0 -A - 5 + C = 0 \quad \Rightarrow \quad C = A + 5 \tag{5}

Substitute equation (4) and (5) into equation (3):

9A+6(1βˆ’A)βˆ’2(A+5)=49A + 6(1 - A) - 2(A + 5) = 4 9A+6βˆ’6Aβˆ’2Aβˆ’10=49A + 6 - 6A - 2A - 10 = 4 (9Aβˆ’6Aβˆ’2A)+(6βˆ’10)=4(9A - 6A - 2A) + (6 - 10) = 4 Aβˆ’4=4β‡’A=8A - 4 = 4 \quad \Rightarrow \quad A = 8

Use A=8A = 8 in equation (4):

B=1βˆ’8=βˆ’7B = 1 - 8 = -7

Use A=8A = 8 in equation (5):

C=8+5=13C = 8 + 5 = 13

Step 6: Write the Partial Fraction Decomposition

Now that we have A=8A = 8, B=βˆ’7B = -7, and C=13C = 13, we can write the partial fraction decomposition as:

x2+4(xβˆ’2)(xβˆ’3)2=8xβˆ’2+βˆ’7xβˆ’3+13(xβˆ’3)2\frac{x^2 + 4}{(x-2)(x-3)^2} = \frac{8}{x-2} + \frac{-7}{x-3} + \frac{13}{(x-3)^2}

This is the final answer using partial fraction decomposition.

Would you like more details or have any questions?

Here are 5 related questions to explore:

  1. How do you determine the form of partial fractions for more complex denominators?
  2. What happens if there is a repeated linear factor like (xβˆ’3)2(x-3)^2?
  3. How do you solve systems of equations with more than three variables?
  4. Can this method be extended to solve higher-order polynomials?
  5. What are the conditions for applying partial fraction decomposition?

Tip: Always compare coefficients carefully when solving for constants in partial fractions to avoid mistakes!

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Math Problem Analysis

Mathematical Concepts

Partial Fraction Decomposition
Algebra
Rational Functions

Formulas

Partial fraction decomposition: A/(x - a) + B/(x - b) + C/(x - b)^2
Solving systems of linear equations
Quadratic expansion: (x - a)^2 = x^2 - 2ax + a^2

Theorems

Method of Partial Fractions
System of Linear Equations

Suitable Grade Level

Grades 10-12

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