Bosco

Find the minimum value of \frac{x^2}{x - 1 + x^3} for x > 1    

I used Desmos to plot y = (x²) / (x – 1 + x³)    

As x increases past (1,1), the value of y decreases asymptotically,    

which means y approaces zero (the x-axis) but it never gets there.    

So there is no minimum value.  We say "it approaches zero".    

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The real numbers x and y satisfy    
x^2 + y^2 - 8x + 6y + 23 = 0.    
Find the largest possible value of x + y.      

Given                               x² + y² – 8x + 6y + 23  =  0     

Group together               (x² – 8x     )  +  (y² + 6y     )  + 23  =  0    

                                       (x² – 8x     )  +  (y² + 6y     )           =  – 23   

Complete both squares and add the same    

amount on the right as you add on the left.    

                                         (x² – 8x + 16) + (y² + 6y + 9)  =  – 23  + 16  + 9     

                                         (x – 4)²  +  (y + 3)²  =  2    

This draws a circle centered at (+4, –3) with a radius the sqrt(2)     

We don't care about the center, all we care about is the radius.    

By formula, the largest x+y of a circle is its radius times the sqrt(2).    

So, the largest x+y is the sqrt(2) times the sqrt(2)         x + y = sqrt(2) • sqrt(2)    

                                                                                       largest x + y = 2    

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Compute    

(2 + 3i) + (2 - 3i) + (-2 + 3i) + (-2 - 3i).     

given                                                           (2 + 3i) + (2 – 3i) + (–2 + 3i) + (–2 – 3i)    

without the parentheses                               2 + 3i + 2 – 3i – 2 + 3i – 2 – 3i    

rearranged for ease of recognition             +2 +2 –2 –2   +3i +3i –3i –3i    

looks like everything cancels out                 zero    

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Give an example of a quadratic function that has zeroes at x = 2 and x = -2, and that takes the value 0 when x = 2.   

zeros at +2 & –2 gives us (x – 2)(x + 2)    

then multiplied out, the quadratic is x² – 4    

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Alex chooses a number at random from the set {1, 2, 3, 4, 5}.  Winnie also chooses a number at random from the same set.  (They can choose the same number.)  What is the probability that the product of their numbers is at least 4?    

There are 25 combinations possible.    

20 of them multiply to 4 or more:     1,4  1,5    

                                                         2,2  2,3  2,4  2,5    

                                                         3,2  3,3  3,4  3,5     

                                                         4,1  4,2  4,3  4,4  4,5     

                                                         5,1  5,2  5.3  5,4  5,5    

Therefore the probability is 20 / 25 or 80%.    

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Evaluate the infinite geometric series    
0.4 + 0.036 + 0.0000324 + . . .     

0.036 / 0.4              =  0.09

0.0000324 / 0.036  =  0.0009    

Definition, from the internet:  In Maths, Geometric Progression is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio.    

The ratios are not the same, so the series in question is not a geometric series.    

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(x+2)/(x)  •  (x+3)/(x+1)  •  (x+4)/(x+2)  =  10    

cancel the (x+2) ' s   

(1)/(x) • (x+3)/(x+1) • (x+4)/(1) = 10    

[ (1) • (x+3) • (x+4) ] / (x) • (x+1) • (1) ] = 10    

(x² + 7x + 12) / (x² + x) = 10    

x² + 7x + 12 = 10x² + 10x    

0 = 9x² + 3x – 12     this will factor    

(3x + 4) • (3x – 3) = 0    

                      3x + 4 = 0     this x will not be an integer, so disregard    

                      3x – 3 = 0    

                            3x = 3    

                              x = 1     therefore (x+2) / (x) = 3/1 = 3 is the value of the first fraction   

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The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?    

x² – mx + 24 = 10     

Subtract 10 from both sides.    

x² – mx + 14 = 0    

m has to be the sum of two negative integers that multiply to 14.    

These could be –1 & –14 or –2 & –7.    

This makes 2 possible values for m; i.e., m could be –15 or –9.    

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A9 multiply by b9 = 2001 .    

Find the value of A and B? A and B are single digit number.     

The factors of 2001 are 1, 3, 23, 29, 69, 87, 667, and 2001.     

Only two factors end with a 9, so they have to be 29 and 69.      

29 • 69 = 2001     A and B are 2 and 6.    

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