geno3141

To factor  12x² - 6x - 90, it would be OK to start by factoring out 3:

--->   12x² - 6x - 90  =  3(4x² - 2x - 30)

This reduces the number of options without removing any possibly correct factors.

Also, if you look at  4x² - 2x - 30  you may notice that each of the coefficients is an even number, meaning that you can now factor out a 2, making the problem simpler.

By factoring out both a 3 and a 2, you are factoring out a 6:

--->   12x² - 6x - 90  =  6(2x² - x - 15)

leaving you with far fewer possibilities.

One general formula for a vertex is:  y - k  =  a(x - h)²     <--->     This parabola has its vertex at (h,k).

If a parabola has a vertex (5,3), its equation can be:  y - 3  =  a(x - 5)²

To find the value of a, place (2,0) into this equation, replacing x with 2 and y with 0:   0 - 3  =  a(2 - 5)²

and solve this for a:     0 - 3  =  a(2 - 5)²     --->     -3  =  a(-3)²     --->     -3  =  a·9     --->    a  =  -1/3

So, the equation is:  y - 3  =  (-1/3)(x - 5)²

Multiplying this out:     y - 3  =  (-1/3)(x² - 10x + 25)

                                   y - 3  =  (-1/3)x² + (10/3)x - (25/3)

                                        y  =  (-1/3)x² + (10/3)x - (16/3)

--->     a = -1/3     b = 10/3     c = -16/3

--->     a + b + c  =  -7/3

Since  log_n(x) = y  can be written as  n^y  =  x  or  x = n^y,

log_n(x)  =  log_a(b)   can be written as   n^log_a(b)  =  x      or     x  =  n^log_a(b)​  

x # y  =  x² / y

     --->     a # 3  =  a² / 3

If  a # 3  =  9     --->     a² / 3  =  9     --->      a²  =  27     --->     a = 3·sqrt(3)     or     a = - 3·sqrt(3)

You have:  (x^1/3)⁴  =  16

Then you took the fourth root of both sides:

      this will give you either  x^1/3  =  2    or    x^1/3  =  -2

Now you need to cube both sides:

      (x^1/3)³  =  (2)³          and         (x^1/3)³  =  (-2)³    

              x  =  8                                    x  =  -8

But, unless you are using complex numbers, logs don't have negative numbers for bases, so the only answer is:  x = 8.

You started correctly, just keep going!

Let   g(x)  =  x⁴ - 5x³ + 2x² + 7x - 11   with roots p, q, r, and s.

Since the roots are p, q, r, and s:   g(x)  =  (x - p)(x - q)(x - r)(x - s)

Multiply out  (x - p)(x - q)(x - r)(x - s)  

     =  x⁴ - x³p - x³​q - x³​r - x³​s + x²pq + x²​pr + x²​ps + x²​qr + x²​qs + x²​rs - xpqr - xpqs - xqrs - xprs - xqrs + pqrs

     =  x⁴ - (p + q + r + s)x³ + (pq + pr + ps + qr + qs + rs)x² - (pqr + pqs + qrs + prs)x + pqrs

Since the coefficients of the x-cubed terms are equal:  - (p + q + r + s)  =  - 5   --->   p + q + r + s  =  5

Since the coefficients of the x-squared terms are equal:  (pq + pr + ps + qr + qs + rs)  =  2

Since the coefficients of the x-terms are equal:  - (pqr + pqs + qrs + prs)  =  + 7   --->   pqr + pqs + qrs + prs  =  -7

and:  pqrs  =  -11

(a)  Compute pqr+pqs+prs+qrs  =  -7

(b)  Compute 1/p+1/q+1/r+1/s:  

       Write with the common denominator of pqrs:  (qrs + prs + pqs + pqr) / pqrs   =   -7/-11  =  7/11

(c)  Compute p^2+q^2+r^2+s^2:

          (p + q + r + s)²  =  p² + q² + r² + s² + 2pq + 2pr + 2ps + 2qr + 2qs + 2rs

                                   =  p² + q² + r² + s² + 2(pq + pr + ps + 2qr + 2qs + 2rs)

                                   =  p² + q² + r² + s² + 2(2)

                                   =  p² + q² + r² + s² + 4

          But, since  p + q + r + s  =  5

                --->     (5)²  =  p² + q² + r² + s² + 4

                --->      25  =  p² + q² + r² + s² + 4

                --->     p² + q² + r² + s²  =  21

(d)  Compute p^2qrs+pq^2rs+pqr^2s+pqrs^2:

          p^2qrs+pq^2rs+pqr^2s+pqrs^2  =  pqrs(p + q + r + s)  =  -11(p + q + r + s)  =  -11(5)  =  -55

If  3 - i⁴sqrt(2)  =  3 - sqrt(2)  because i⁴ = 1.

If  3 - sqrt(2) is a root then (if the coefficients are rational), 3 + sqrt(2) is also a root.

If it's a quartic, you will have 2 more roots. They will be two rational roots [such as 2/3 and -7], or a pair of conjugate complex roots [such as 2 + 3i and 2 - 3i], or a pair of conjugate irrational roots [such as -5 + sqrt(19) and -5 - sqrt(19)].

(x - (3 - sqrt(2) ) · (x - (3 + sqrt(2)) · ( x - (something) ) · ( x - (something)  =  0

You can change any base into base 10 by using this formula:  log_b(x) = log₁₀(x) / log₁₀(b).

For instance, you can find log₇(114) by using log₁₀(114) / log₁₀(7).

You need not change the log into base 10; it can be changed into any base.

If you want to use ln, log_b(x) = ln(x) / ln(b).

Example, you can find log₇(114) by uing ln(114) / ln(7).

Your question isn't clear; you entered Q((1 + sqrt(2)⁸) = 0.

There are 3 left-parantheses but only 2 right-parantheses.

I'm going to change the expression to:  Q( ( 1 + sqrt(2) )⁸ )  =  0.

If  Q(x)  =  x² + bx + c:

( 1 + sqrt(2) )⁸  =  577 + 408sqrt(2)

--->   Q( ( 1 + sqrt(2) )⁸ )  =  Q( 577 + 408sqrt(2) )  =  ( 577 + 408sqrt(2) )² + ( 577 + 408sqrt(2) )b + c  =  0

( 577 + 408sqrt(2) )²  =  665857 + 470832sqrt(2)

--->   ( 577 + 408sqrt(2) )² + ( 577 + 408sqrt(2) )b + c  =  0    

--->     665857 + 470832sqrt(2) + 577b + 408sqrt(2)b + c  =  0

Separate this into two equations, one have square roots, the other without:

--->     665857 + 577b + c  =  0     and     470832sqrt(2) + 408sqrt(2)b  =  0

                                                                                           408sqrt(2)b  =  -470832sqrt(2)

                                                                                                           b  =  -1154

When  b = -1154:     665857 + 577b + c  =  0

                       665857 + 577(-1154) + c  =  0

                             665857 - 665858 + c  =  0

                                                    -1 + c  =  0

                                                           c  =  1

b + c  =  -1154 + 1  =  -1153

If  f(x)  =  x³ + bx + c                (This isn't the problem, but I don't understand the problem as written.)

If  f( 5 + sqrt(3) )  =  0,  determine  b + c.

Replace  5 + sqrt(3)  into  f(x)  =  x³ + bx + c     --->     f( 5 + sqrt(3) )  =  ( 5 + sqrt(3) )³ + b( 5 + sqrt(3) ) + c  =  0

Since  ( 5 + sqrt(3) )³   =  170 + 78sqrt(3)     --->                                      170 + 78sqrt(3) + 5b + sqrt(3)b + c  =  0

To have integer coefficients, the sqrt(3) terms must cancel, so create two smaller equations, one involving only the sqrt terms and other involving only the integer terms:

--->   170 + 5b + c  =  0     and     78sqrt(3) + sqrt(3)b  =  0

                                                                       sqrt(3)b  =  -78sqrt(3)

                                                                                 b  =  -78

If b = -78:     170 + 5b + c  =  0

                     170 + 5(-78) + c  =  0

                     170 - 390 + c  =  0

                            -220 + c  =  0

                                       c  =  220

b + c  =  -78 + 220  =  142