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 #1
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Absolutely, I’ve been improving my problem-solving abilities in solving polynomial equations. Let's find a3+b3, where a and b are the roots of the equation: 5x2−11x+4=−3x2−17x+5

We can solve the equation for a and b using the quadratic formula and then use the fact that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​.

Steps to solve: 1. Solve the equation for a and b: 5x2−11x+4=−3x2−17x+5

Combining like terms and rearranging the equation, we get: 8x2+6x−1=0

Using the quadratic formula, we get: x=2a−b±b2−4ac​​ where a, b, and c are the coefficients of the quadratic equation. In this case, a = 8, b = 6, and c = -1. Substituting these values into the formula, we get:

x=2⋅8−6±62−4⋅8⋅−1​​

x=16−6±217​​

Therefore, the roots are x=817​−3​ and x=8−17​−3​.

2. Find a^3 + b^3: We know that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​. In this case, we have:

a+b=−86​=−43​ ab=8−1​

We can use these relationships to find a3+b3:

a3+b3=(a+b)(a2−ab+b2)

Substitute the values we found for a + b and ab:

a3+b3=−43​(a2−ab+b2)

We don't need to find the individual values of a^2 and b^2$ since we can rewrite the expression using the fact (a+b)2=a2+2ab+b2 :

(a+b)2=a2+2ab+b2

Substitute a + b = -3/4:

(−43​)2=a2+2ab+b2

Expand:

169​=a2+2ab+b2

Substitute ab = -1/8:

169​=a2−41​+b2

Combine like terms:

169​+41​=a2+b2

a2+b2=45​

Substitute this back into the expression for a^3 + b^3:

a3+b3=−43​(45​−81​)

a3+b3=−43​⋅89​

a3+b3=−3227​

Answer: a3+b3=−3227​

2 avr. 2024
 #1
avatar+1320 
0

There are two cases to consider for placing the two indistinguishable pieces on the chessboard:

 

Same Row: The pieces can be placed on any of the 8 rows. Once the row is chosen, either piece can be placed on any of the 8 squares in that row.

 

So, for the same row scenario, there are 8 ways to pick a row and 8 ways to arrange the pieces within that row, resulting in 8 * 8 = 64 arrangements.

 

Same Column (excluding diagonal): The pieces can be placed on any of the 8 columns (excluding the possibility of them being on the same diagonal, which we'll address later).

 

Similar to the same row case, once a column is chosen, either piece can be placed on any of the 8 squares in that column.

 

However, this counts the diagonal placement twice (once for each way of choosing the row or column first).

 

So, for the same column scenario (excluding diagonal), there are 7 ways to pick a column and 7 ways to arrange the pieces within that column, resulting in 7 * 7 = 49 arrangements.

 

Correcting for Diagonal Placement:

 

We've double-counted the arrangements where the pieces are on the same diagonal.

 

How many are there? Imagine choosing the diagonal first. There are only 4 choices (one for each corner).

 

Then, you can place the pieces on two squares out of the 8 on that diagonal, giving 8 choices.

 

However, since we've already counted this scenario twice (once for each way of choosing the row or column first), we need to subtract 1.

 

Therefore, the number of diagonal placements counted twice is 4 (diagonals) * (8C2) (combinations to choose 2 squares out of 8) - 1 (double counting) = 14.

 

Total Arrangements:

 

Adding the arrangements for same row, same column (excluding diagonal), and subtracting the double-counted diagonals, we get the total number of possible placements:

 

Total = Same Row + Same Column (excluding diagonal) - Double Counted Diagonals Total = 64 + 49 - 14 Total = 99

 

There are 99 ways to place two indistinguishable pieces on an 8x8 chessboard if they must be in the same row or the same column.

2 avr. 2024
 #2
avatar+1320 
+1
24 mars 2024