geno3141

Find the integer values of b and c so that the polynomial  x² - x - 1  and the polynomial  x⁵ - bx - c have the same roots.

First:  find the roots of  x² - x - 1  =  0    

--->     Using the quadratic formula, the roots are  [ 1 + sqrt(5) ] / 2   and   [ 1 + sqrt(5) ] / 2.

Placing these values into the equation  x⁵ - bx - c  =  0,  we get:

x  =  [ 1 + sqrt(5) ] / 2   --->   ( [ 1 + sqrt(5) ] / 2 )⁵   -  b( [ 1 + sqrt(5) ] / 2 )  -  c  =  0

                                    --->    [ 176 + 80sqrt(5) ] / 32  -  ( [ 1 + sqrt(5) ] / 2 )b  -  c  =  0

    multiplying by 32     --->    176  +  80sqrt(5)  -  16b  -  16sqrt(5)b  -  32c  =  0

Using the same steps for x  =  [ 1 - sqrt(5) ] / 2 , we get:     176  -  80sqrt(5)  -  16b  +  16sqrt(5)b  -  32c  =  0 

Now, let's combine these two results:      176  +  80sqrt(5)  -  16b  -  16sqrt(5)b  -  32c  =  0

                                                                 176  -  80sqrt(5)  -  16b  +  16sqrt(5)b  -  32c  =  0 

Subtract the lower equation from the upper:      160sqrt(5)  -  32sqrt(5)b  =  0

     --->     160sqrt(5)  =  32sqrt(5)b     --->     b  =  5

Now, substitute this value for b into  176  +  80sqrt(5)  -  16b  -  16sqrt(5)b  -  32c  =  0

     --->                                          176  +  80sqrt(5)  -  16(5)  -  16sqrt(5)(5)  -  32c  =  0    

     --->                                                    176  +  80sqrt(5)   - 80  -  80sqrt(5)  - 32c  =  0

     --->                                                                                                      96  -  32c  =  0

     --->                                                                                                                   c  =  3                    

y  =  x² - 3x + 2

To find the roots (a,0) and (b,0), let y = 0:

y  =  x² - 3x + 2      --->     0  =  x² - 3x + 2      --->     0  = (x - 2)(x - 1)

                                                                       --->          x = 2  or  x = 1     --->     (2,0) and (1, 0)

The point where the graph crosses the y-axis has an x-value of 0:

y  =  0² - 3(0) + 2     --->     y = 2      --->     (0,2)

a = 1 and b = 2 and c = 2     or     a = 2 and b = 1 and c = 2

By writing the function as  P(x)  =  (x - 3)² + 1,  you are showing that the minimum y-value of the function is 1.

[Since (x - 3)² has either a value of 0 or a positive value for any choice of x, the graph rises from the minimum value, which occurs when the value of (x - 3)² = 0. When the "+1" is added, the minimum y-value must be 1.]

Since the minimum y-value must be 1, choice 1) is incorrect because zeroes occur only when the y-value is 0.

Choice 2) is correct.

Choice 3) is incorrect because the minimum is 1, not 3.

Choice 4) is incorrect because there are no zeroes.

To find the largest power of 2 that divides into (2⁴)!:

First:  simplify (2⁴)!   --->   (2⁴)!  =  (16)!

16! = 16 x 15 x 14 x ... x 2 x 1

But, if we want to find a number that divides into 16!, we don't really care about the odd numbers in  16 x 15 x ... x 2 x 1.

All we need to look at are the even factors:  16 x 14 x 12 x 10 x 8 x 4 x 2.

16  = 2⁴

14 = 2 x 7

12 = 2² x 3

10 = 2 x 5

8 = 2³

6 = 2 x 3

4 = 2²

2 = 2

So, the total number of factors is 4 + 1 + 2 + 1 + 3 + 1 + 2 + 1  =  15     <---   Answer

As a check:  16! / 2¹⁵  =  638 512 875  (no further factors of 2)

log₁₀(log₁₀N) = 1     --->     log₁₀N = 10¹     --->     log₁₀N = 10     --->     N = 10¹⁰

loge(logeN) = 1     --->     logeN = e¹     --->     logeN = e     --->     N = e^e    

If two triangles are similar, the ratio of their areas equals the ratio of the squares of their corresponding sides.

     Area₁ / Area₂  =  side₁² / side₂²

--->   270 / x  =  25² / 35²

--->   270 / x  =  625 / 1225          Cross multiply:

--->   1225·270  =  625·x

--->   330 750  =  625x                 Divide by 625:

--->     x  =  529.2

Notice that the sides are squared but the areas aren't.

Let me try part d     ---     but I'm guessing at what you mean ....

i)     f(x)  =  2 + x³(7·ln(x) - 9)          Using the product rule:

       f'(x)  =  2 + x³(7/x) + 3x²(7·ln(x) - 9)

       f'(x)  =  2 + 7x² + 3x²(7·ln(x) - 9)

       f'(1)  =  2 + 7·1² + 3·1²(7·ln(1) - 9

              --->   f'(1)  =  2 + 7 + 3(7·0 - 9)   =   2 + 7 - 27   =  -18

ii)    g(x)  =  1 + 2x²e^x + x²                        [Is this right?]

       g'(x)  =  0 + 2x²e^x + 4x·e^x + 2x

       g'(x)  =   2x²e^x + 4x·e^x + 2x

              --->   g'(1)  =  2·1²·e¹ + 4·1·e¹ + 2·1  =  2e + 4e + 2  =  6e + 2

iii)   h(x)  =  sqrt(4) + x⁴                 [Is this right?]

       h'(x)  =  0 + 4x³

       h'(x)  =  4x³

              --->   h'(1)  =  4·1³  =  4

Let me try part e:     f(x)  =  x³ - 2x² - 4x + 5

i)     f'(x)  =  3x² - 4x - 4

       f''(x)  =  6x - 4

ii)     The critical points occur where the first derivate either does not exist or it equals 0.

        Since the first derivative is always defined, we need to find where it equals 0.

               f'(x)  =  3x² - 4x - 4  =  0     --->     3x² - 4x - 4  =  0     --->     (3x + 2)(x - 2)  =  0

                                                         --->      x  =  -2/3     or     x = 2           <---     the critical points

iii)     A local maximum occurs where the first derivative equals 0 and the second derivative is less than 0.

              f'(-2/3)  =  0

              f''(-2/3)  =  6(-2/3) - 4  =  -8         <---     Therefore, a local maximum occurs where  x = -2/3.

iv)     A local minimum occurs where the first derivative equals 0 and the second derivative is greater than 0.

              f'(2)  =  0

              f''(2)  =  6(2) - 4  =  8          <--     Therefore, a local minimum occurs where x = 2.

The solution to the system of equations:

     y  =  1.5x - 4

     y  =  -4

Replace the y-term in the first equation with -4:

     -4  =  1.5x - 4           Add 4 to both sides:

      0  =  1.5x                 Divide both sides by 1.5:

       0  =  x

The solution is:  x = 0,  y = -4

y  =  (x - 5)(x - 2)³

If the factor has an odd power, it will pass through that point; therefore the equation passes through the points  x = 5  and  x = 2.  

The graph will go directly through the point if the power is 1, it "wiggles through" if the power is 3, 5, 7 etc.

If the power is an even number, it touches that point and bounces back to the direction from which it came.