Q. Find the values of x.4x+14x2−3x−1+x2−x+1x3+1=2
Identify Terms and Denominator: First, let's identify the individual terms of the equation and understand that we need to find a common denominator to combine the fractions on the left side of the equation.
Find Common Denominator: The common denominator for the two fractions (4x+1) and (x2−x+1) would be their product, (4x+1)(x2−x+1). We will multiply both the numerator and denominator of each fraction by the necessary terms to achieve this common denominator.
Multiply First Fraction: Multiply the first fraction (4x2−3x−1)/(4x+1) by (x2−x+1)/(x2−x+1) to get ((4x2−3x−1)(x2−x+1))/((4x+1)(x2−x+1)).
Multiply Second Fraction: Multiply the second fraction (x3+1)/(x2−x+1) by (4x+1)/(4x+1) to get ((x3+1)(4x+1))/((4x+1)(x2−x+1)).
Combine Fractions: Now, we will add the two fractions together since they have the same denominator. This gives us: (4x+1)(x2−x+1)(4x2−3x−1)(x2−x+1)+(x3+1)(4x+1)=2
Expand First Numerator: We need to expand the numerators of both fractions before adding them. Let's start with the first numerator (4x2−3x−1)(x2−x+1). We will use the distributive property (FOIL) to expand this product.
Expand Second Numerator: Expanding (4x2−3x−1)(x2−x+1) gives us 4x4−4x3+4x2−3x3+3x2−3x−x2+x−1.
Add Expanded Numerators: Combining like terms in the expanded expression gives us 4x4−7x3+6x2−3x−1.
Combine Like Terms: Now let's expand the second numerator (x3+1)(4x+1) using the distributive property.
Eliminate Denominator: Expanding (x3+1)(4x+1) gives us 4x4+x3+4x+1.
Expand Right Side: Now we add the expanded numerators together: 4x4−7x3+6x2−3x−1 + 4x4+x3+4x+1.
Combine Like Terms Right Side: Combining like terms gives us 8x4−6x3+6x2+x.
Distribute Right Side: Now we have the combined fraction as (8x4−6x3+6x2+x)/((4x+1)(x2−x+1))=2.
Set Equation to Zero: To solve for x, we need to get rid of the denominator. We can do this by multiplying both sides of the equation by the denominator, which gives us 8x4−6x3+6x2+x=2(4x+1)(x2−x+1).
Subtract Right Side: Now we need to expand the right side of the equation: 2(4x+1)(x2−x+1).
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1).
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1).Combining like terms on the right side gives us 2(4x3−3x2+3x+1).
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1). Combining like terms on the right side gives us 2(4x3−3x2+3x+1). Distributing the 2 across the terms on the right side gives us 8x3−6x2+6x+2.
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1). Combining like terms on the right side gives us 2(4x3−3x2+3x+1). Distributing the 2 across the terms on the right side gives us 8x3−6x2+6x+2. Now we have the equation 8x4−6x3+6x2+x=8x3−6x2+6x+2.
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1). Combining like terms on the right side gives us 2(4x3−3x2+3x+1). Distributing the 2 across the terms on the right side gives us 8x3−6x2+6x+2. Now we have the equation 8x4−6x3+6x2+x=8x3−6x2+6x+2. We need to bring all terms to one side to set the equation to zero: 8x4−6x3+6x2+x−(8x3−6x2+6x+2)=0.
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1). Combining like terms on the right side gives us 2(4x3−3x2+3x+1). Distributing the 2 across the terms on the right side gives us 8x3−6x2+6x+2. Now we have the equation 8x4−6x3+6x2+x=8x3−6x2+6x+2. We need to bring all terms to one side to set the equation to zero: 8x4−6x3+6x2+x−(8x3−6x2+6x+2)=0. Subtracting the terms on the right from the left gives us 8x4−14x3+12x2−5x−2=0.
Quartic Equation Error: Expanding the right side gives us 2(4x3−4x2+4x+x2−x+1). Combining like terms on the right side gives us 2(4x3−3x2+3x+1). Distributing the 2 across the terms on the right side gives us 8x3−6x2+6x+2. Now we have the equation 8x4−6x3+6x2+x=8x3−6x2+6x+2. We need to bring all terms to one side to set the equation to zero: 8x4−6x3+6x2+x−(8x3−6x2+6x+2)=0. Subtracting the terms on the right from the left gives us 8x4−14x3+12x2−5x−2=0. This is a quartic equation, and solving it algebraically can be very complex. We would typically use numerical methods or graphing to find the roots. However, there is a mistake in the previous steps. The derivative of ln(2x) is not 1/x as stated in Step 3 of the full solution. The correct derivative should be 1/x multiplied by the derivative of the inner function 2(4x3−3x2+3x+1)0, which is 2. Therefore, the correct derivative of ln(2x) is 2(4x3−3x2+3x+1)3. This means there is a math error in the full solution provided.
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