Guide :

# simplify each sum or difference with operations with radical expressions

6squareroot8 - 2squareroot50

## Research, Knowledge and Information :

### Algebra Examples | Radical Expressions and Equations ...

Radical Expressions and Equations. Simplify. Rewrite as . Pull terms out from under the radical, assuming positive real numbers.
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### 12.2 Operations with Radical Expressions

Operations with Radical Expressions ... Use sum and difference pattern. ... To simplify some radical expressions with radicals in the denominator, ...
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### Simplifying Radical Expressions Calculator

Simplifying Radical Expressions Calculator. Important: The form will NOT let you enter wrong characters (like y, p, ;, ...) How to input??
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### 8.3 Operations on Radical Expressions

8.3 Operations on Radical Expressions ... sum as or The LCD of the ... Simplify each expression. (a) (b) (c) 5 13 9 13a 15b 5 17
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### Alg 10 3

Operations with Radical Expressions Simplify each sum or difference. ... 10-3 Operations with Radical Expressions ... Simplify each expression.
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### simplify the sum calculator - Polymathlove.com

In the event you actually require assistance with math and in particular with simplify the sum ... Operations on Radicals: Factoring a Difference ... radical ...
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### Simplifying Expressions Calculator | Wyzant Resources

Simplifying Expressions Calculator. ... radicals, and fractions are not supported. Tutoring. ... Order of Operations; Simplifying Negative Exponents;
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### Operations with Radical Expressions - K Rohlwing

Operations with Radical Expressions Simplify each sum or difference. Simplify each quotient. ... Error Analysis A student multiplied the radical expressions shown at the
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### Radical Equation Calculator - Symbolab

Radical Equation Calculator Solve radical equations, step-by-step. ... Basic Operations. Simplify; Factor; Expand; Join; ... \sum \sum _{n=0}^N
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## Suggested Questions And Answer :

### With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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### how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61
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### simplify each sum or difference with operations with radical expressions

start: 6root(8) -2root(50) root(8)=2 root(2) root(50)=root(2*25)=5root(2) 6root(8)-->12*root(2) 2root(50)-->10*root(2) differ=(12-10)root(2) =2root(2)
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### what is the nth term of 5,14,31,56,89,130,179,236

If we calculate the difference between one number and the next we get: 9 17 25 33 41 ... Now the difference between each of these differences is 8, suggesting a fixed pattern involving 8. We can call the first term in the series a0=5, the next a1 and so on until a(n). We can see that the differences are related to 8 by position: 1+0*8=1; 1+1*8=9; 1+2*8=17; 1+3*8=25; 1+4*8=33, etc. so for the nth term 1+8n gives us the difference from the first term, which is 5. The series becomes 5, 5+8*1+1, 5+8*1+1+8*2+1, 5+8*1+1+8*2+1+8*3+1, ... We should now be able to work out the nth term. We start with 5 and add n to account for the progression of 1's. Then we have another arithmetic progression multiplied by 8: 8(1+2+3+4...n). The sum to n terms of the natural numbers is n(n+1)/2 so multiplying by 8 we get 4n(n+1). Therefore the expression for the nth term is 5+n+4n(n+1). Let's try it for n=5 and we get 130, which matches the given sequence. The expression can be simplified: 5+n(1+4n+4)=5+n(4n+5). When n=0, the term is 5; when n=1 it's 14. When n=7 the term is 236.
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### how do you simplify an expression

one set of parentheses to write an expression that simplifies to 7, 13, or 17. Do your work step by step and explain each step as you simplify the expression. Demonstrate the consequences of not using the proper order of operations by showing that other orders yield different answers.
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### sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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### Probability Question

There are 70 combinations of quartets of these 8 numbers, or, in other words, 70 different ways of grouping 4 outcomes from S. They all have the same probability of being selected as a group because each member of S has the same probability of selection. These are combinations of the possible outcomes of simply picking 4 numbers from 1 to 8 so that once drawn, a number cannot be returned to the pool. After a group of 4 has been drawn, then all the numbers are returned to the pool for the next group selection in this experiment. If we add together the numbers in each quartet we get a particular sum, associating each quartet with a sum. For example, 1+2+3+4=10 and 5+6+7+8=26. So the sum is in the range 10 to 26, and the sums are not usually unique. For example: 1+2+3+8=1+3+4+6=2+3+4+5=14. So, given a sum we can identify how many quartets have that sum. Or, we can find out how many have less or more than this sum. That gives us a probability based on the number of occurrences divided by the total number of possibilities. If we can find a sum that is shared by 49 quartets then the probability of selecting that sum will be 49/70=7/10=0.7. Alternatively, we can count the number of quartets that are below or above that sum. This will be one of the requirements of the experiment. But it doesn't have to be a sum. We can use other operators than plus. In fact, between 4 numbers we can have three operators, for example, 1*2+3*4=14, or 1*2*3*4=24. So two possible events for X will simply be the result of applying different operators. If we take the sample space for the operators as { + - * } then there will be 6 permutations of these, if all operators are to be applied in order, to ordered elements of each quartet. The elements can be ordered in ascending or descending order and the results will usually be different. The normal priority rules would apply (multiplication takes priority over addition and subtraction). All that's needed is a particular result (< X, ≤ X, = X, ≥ X, > X), so that the number of occurrences is 49 out of the possible 70 outcomes. Then P(X)=49/70=0.7. The question doesn't ask for specific X, just possibilities for X. If I can, I'll identify specific events for X...when I have time!
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### what is this pattern? 2, 5, 9, 19, 40, 77, 137

Compound Arithmetic Sequence a0          a1          a2          a3          a4          a5          a6         b0          b1         b2          b3          b4           b5  ------- 1st differences               c0          c1          c2          c3          c4  -------------- 2nd differences                        d            d            d             d   ------------------- 3rd differences = constant We have here an irregular sequence (a_n) = (a_1, a_2, a_3, ..., a_k, ...). The differences between the elements of (a_n), the 1st differences, are non-constant. So also are the differences between the elements of (b_n), the 2nd differences. However, the differences between the elements of (c_n) are constant. This makes (c_n) a regular arithmetic sequence, and so we can write, cn = c0 + nd,  n = 0,1,2,3,... The sequence (b_n) is non-regular, but we can write, b_(n+1) = b_n + c_n Using the expression for the c-sequence, b_(n+1) = b_n + c_0 + nd Solve the recurrence relation for b_n Develop the terms of the sequence. b1 = b0 + c0 b2 = b1 + c0 + 1.d = b0 + 2.c0 + 1.d b3 = b2 + c0 + 2.d = b0 + 3.c0 + (1 + 2).d b4 = b3 + c0 + 3.d = b0 + 4.c0 + (1 + 2 + 3).d ‘ ‘ ‘ b(n+1) = b_0 + (n+1).c0 + sum[k=1..n](k) * d b_(n+1) = b_0 + (n+1).c0 + (nd/2)(n + 1) b_(n+1) = b_0 + (n+1)(c_0 + nd/2) We could also write c0 = b1 – b0 in the above expression. Now that we have a rule/expression for the general term of the b-sequence, let us analyse the a-sequence. Since the sequence (a_n) is non-regular like (b_n), we can write, a_(n+1) = a_n + b_n Substituting for the general term of the b-sequence, a_(n+1) = a_n + b0 + n(c0 + (n-1)d/2)      (co can be replaced later with c0 = b1 – b0) Solve the recurrence relation for a_n Develop the terms of the sequence. a1 = a0 + b0 a2 = a1 + b0 + 1.c0 + 1.0.d/2 = a0 + 2.b0 + 1.c0 + 1.0.d/2 a3 = a2 + b0 + 2.c0 + 2.1.d/2 = a0 + 3.b0 + (1 + 2).c0 + (1.0 + 2.1).d/2   (N.B. here, 1.0 = 1*0, 2.1 = 2*1) a4 = a3 + b0 + 3.c0 + 3.2.d/2 = a0 + 4.b0 + (1 + 2 + 3).c0 + (1.0 + 2.1 + 3.2).d/2 . . . a_n = a0 + n.b0 + sum[k=1..n-1] k * c0 + sum[k=1..n-1](k(k-1)) * (d/2) a_n = a0 + n.b0 + (c0/2).n(n-1) + sum[k=1..n-1](k^2) * (d/2) – sum[k=1..n-1](k) * (d/2) a_n = a0 + n.b0 + (c0.n/2)(n-1) + (d/2){(n/6)(n-1)(2n-1) – (n/2)(n-1)} a_n = a0 + n.b0 + (c0.n/2)(n-1) + (nd/6)(n^2 – 3n + 2) Substituting for c0 =  b1 – b0, b1 = a2 – a1, b0 = a1 – a0 Then, c0 = a0 – 2a1 + a2. These substitutions give, a_n = (1 + (n/2)(n – 3)).a0 – n(n – 2).a1 – (n/2)(1 – n).a2 + (nd/6)(n^2 – 3n + 2) Now substituting for a0 = 2, a1 = 5, a2 = 9 and d = 5, the rule is, a_n = 2 + n(n – 3) – 5n(n – 2) – (9/2).n.(1 – n) + (5/6).n.(n^2 – 3n + 2)
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### a wire of length l is cut into two parts. One part is bent into a circle and the other into square

If the radius of the circle is d then its area is (pi)d^2 and its circumference is 2(pi)d, the length of wire we need to make the circle. The length of the remainder of the wire is l-2(pi)d, out of which we make the square. So the side of the square is a quarter of this perimeter, 1/4(l-2(pi)d), and the area of the square is the square of this side, 1/16(l-2(pi)d)^2. The sum of the areas of the circle and square is (pi)d^2+1/16(l-2(pi)d)^2. We need the minimum value of this expression, where the variable is d. So we differentiate it with respect to d. That's the same as getting the difference quotient. The expansion of 1/16(l-2(pi)d)^2 is l^2/16-1/4(pi)ld+1/4(pi)^2d^2. [Please distinguish between 1 and l in the following.] The difference quotient is zero at a maximum or minimum, so we have 2(pi)d-l/4(pi)+1/2(pi)^2d=0. We can take out (pi), leaving 2d-l/4+1/2(pi)d=0. Multiply through by 4 to get rid of the fractions: 8d-l+2(pi)d=0, from which d=l/(8+2(pi)). Half the length of the side of the square is 1/8(l-2(pi)d). If we substitute for d in this expression we get 1/8(l-2(pi)l/(8+2(pi)))  = l/8((8+2(pi)-2(pi))/(8+2(pi)) = l/8(8/(8+2(pi)) = l/(8+2(pi)) = d (QED). Therefore the radius of the circle = half the length of the side of the square is either a maximum or minimum value of the expression for the sum of the areas of the circle and square. We can see that this expression gets bigger as d gets bigger, because (pi)d^2 has a positive value always, so we do indeed have a minimum rather than a maximum. We can substitute d=l/(2(4+(pi))) in the expression for the sum of the areas and we get the minimum: (pi)d^2+1/16(l-2(pi)d)^2 = (pi)l^2/4(4+(pi))^2 + 1/16(l-2(pi)l/(2(4+(pi)))^2 = (pi)l^2/4(4+(pi))^2 + l^2/16(1-(pi)/(4+(pi))^2 = (pi)l^2/4(4+(pi))^2 + l^2/16(4+(pi)-(pi))^2(4+(pi))^2 = (pi)l^2/4(4+(pi))^2 + l^2/(4+(pi))^2 = l^2/(4(4+(pi))) or (l^2/4)*1/(4+(pi)) Sorry, it was getting difficult to represent the expressions using this tablet so I've had to accelerate the last bit! I hope I didn't make any mistakes!
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### Find m if x – 3 and x + 2 are factors of x3 + m2x2 – 11x – 15m

36r4 + 36r3 + 3r2) / 9r3  this is normally written as (36r^4 + 36r^3 + 3r^2) / (9r^3)  this is equivalent to (36r^4 / 9r^3) + (36r^3 / 9r^3) + (3r^2 / 9r^3)  the rule for exponent arithmetic to apply is that x^a/x^b = x^(a-b).  so, basically, if the base is the same, then you subtract the exponent in the denominator from the exponent in the numerator and put the result in the numerator.  the other rule for exponent arithmetic to apply is that x^-a = 1/x^a.  you take the reciprocal and change the sign of the exponent.  x^-a = 1/x^a  x^a = 1/x^-a  normally you want to simplify by making all the exponents positive, so if you have x^-a, then show it as 1/x^a, and if you have 1/x^-a, then show it as x^a.  looking at each of the parts of (36r^4 / 9r^3) + (36r^3 / 9r^3) + (3r^2 / 9r^3) individually, you get:  36r^4 / 9r^3 becomes 36/9 * r^4/r^3 which becomes 4 * r^(4-1) which becomes 4r^1 which becomes 4r.  36r^3 / 9r^3 becomes 36/9 * r^3/r^3 which becomes 4 * r^(3-3) which becomes 4 * r^0 which becomes 4 * 1 which becomes 4.  3r^2/9r^3 becomes 3/9 * r^2/r^3 which becomes 1/3 * r^(2-3) which becomes 1/3 * r^-1 which becomes 1/3 * 1/r^1 which becomes 1/3 * 1/r which becomes 1/(3r)  your simplified expression becomes 4r + 4 + 1/(3r)  this should be your solution, but if you wanted to put everything under the same denominator, then you would get:  4r + 4 + 1/(3r) becomes (4r*3r + 4*3r + 1) / (3r)  simplify this to get:  (12r^2 + 12r + 1) / (3r)  that should be your simplified expression.  if we did this correctly, your original expression and your simplified expression should give you the same answer if you replace r with a random value.  for example, assume r = 7  your original expression of (36r^4 + 36r^3 + 3r^2) / (9r^3) becomes equal to 32.04761905  your simplified expression of (12r^2 + 12r + 1) / (3r) becomes equal to 32.04761905  since they both provide the same answer, they're equivalent and you can reasonable conclude that you simplified correctly.  if your stated solution is not this, then let me know what it is and i'll come up with a reason why it should be that and not what i just told you.  the difference, if any, is usually in what the definition of simplified is.
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