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Answer:

Here's a more formal way to introduce #i# than is normally done at school...

Explanation:

The imaginary unit #i# is usually introduced as "the" square root of #-1#, but note that #-1# has two square roots. So how do we tell which is which?

A more formal way to do things is to define the arithmetic of complex numbers in terms of arithmetic on pairs of real numbers, i.e. elements of #RR^2# ...

For any real numbers #a, b, c, d#, define addition and multiplication as follows:

#(a, b) + (c, d) = (a+c, b+d)#

#(a, b) * (c, d) = (ac-bd, ad+bc)#

Then we find that:

  • There is an identity (i.e. zero) for addition:

    #(a, b) + (0, 0) = (0, 0) + (a, b) = (a, b)#

  • Every element has an additive inverse:

    #(a, b) + (-a, -b) = (-a, -b) + (a, b) = (0, 0)#

  • Addition is associative:

    #((a, b) + (c, d)) + (e, f) = (a, b) + ((c, d) + (e, f))#

  • Addition is commutative:

    #(a, b) + (c, d) = (c, d) + (a, b)#

  • There is an identity (i.e. one) for multiplication:

    #(a, b) * (1, 0) = (1, 0) * (a, b) = (a, b)#

  • Every non-zero element has a multiplicative inverse:

    #(a, b) * (a/(a^2+b^2), -b/(a^2+b^2)) = (a/(a^2+b^2), -b/(a^2+b^2)) * (a, b) = (1, 0)#

  • Multiplication is associative:

    #((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d) * (e, f))#

  • Multiplication is commutative:

    #(a, b) * (c, d) = (c, d) * (a, b)#

  • Multiplication is left and right distributive over addition:

#(a, b) * ((c, d) + (e, f)) = ((a, b) * (c, d)) + ((a, b) * (e, f))#

#((a, b) + (c, d)) * (e, f) = ((a, b) * (e, f)) + ((c, d) * (e, f))#

In summary: the elements of #RR^2# form a field under the operations of addition and multiplication defined.

  • The function from #RR# to #RR^2# defined by:

    #f(x) = (x, 0)#

preserves addition, multiplication, etc. so is a field homomorphism.

So we can say that this function embeds the real numbers in the complex numbers - for that is the field that we have defined.

Effectively we can say that #(a, 0)# is a real number and #(a, b)# is a non-real complex number if #b != 0#.

We can then change notation and write our complex numbers as:

#a+bi#

where #a, b in RR# and #i = (0, 1)# is the imaginary unit.

There's our definition:

#i = (0, 1)#

Answer:

Here's the trigonometric method for the case where the cubic equation has #3# real roots...

Explanation:

Let us deal with the reduced cubic:

#t^3+pt+q = 0#

(see other solution for how to get here if your cubic is more general)

Consider the substitution:

#t = k cos theta#

Then our equation becomes:

#k^3 cos^3 theta + kp cos theta + q = 0#

We want to use the trigonometric identity:

#cos 3 theta = 4 cos^3 theta - 3 cos theta#

So dividing our equation by #k^3/4# we find:

#4 cos^3 theta + (4p)/k^2 cos theta + (4q)/k^3 = 0#

Then we want:

#(4p)/k^2 = -3#

So:

#k^2 = -(4p)/3#

and we can choose:

#k = sqrt(-(4p)/3) = 2/3sqrt(-3p)#

Note also:

#k^4 = (-(4p)/3)^2 = (16p)/9#

Then:

#(4q)/k^3 = (4qk)/k^4 = (4qk)/((16p)/9) = (9qk)/4 = 3/2qsqrt(-3p)#

So our equation becomes:

#4 cos^3 theta-3 cos theta + 3/2qsqrt(-3p) = 0#

That is:

#cos 3 theta = -3/2qsqrt(-3p)#

So:

#3 theta = +-cos^(-1)(-3/2qsqrt(-3p)) + 2npi" "n in ZZ#

Hence distinct values:

#cos theta = cos(1/3 cos^(-1)(-3/2qsqrt(-3p))+(2npi)/3)" "n = 0, 1, 2#

Then #t = k cos theta#, so the distinct roots of our original cubic are:

#t_n = 2/3sqrt(-3p) cos(1/3 cos^(-1)(-3/2qsqrt(-3p))+(2npi)/3)" "n = 0, 1, 2#

Footnote

If not all of the roots are real, then #abs(-3/2qsqrt(-3p)) > 1#. The formula we derived is still true, but we have to deal with complex values of #cos^(-1)#.

Alternatively, note that #cosh theta# also satisfies:

#cosh 3 theta = 4 cosh^3 theta - 3 cosh theta#

So we can follow a similar derivation to find:

#t_n = 2/3sqrt(-3p) cosh(1/3 cosh^(-1)(-3/2qsqrt(-3p))+(2npi)/3i)" "n = 0, 1, 2#

I have not really used this hyperbolic method, since in the cases where it makes most sense we can use Cardano's method anyway.

Answer:

Almost. If #k# is any constant then #f(x) = k e^x# satisfies #f'(x) = f(x)#.

Uniqueness of #e^x# is given by requiring #f(x)# to be defined everywhere, and #f(0) = 1#.

Explanation:

Any function of the form #f(x) = ke^x# satisfies #f'(x) = f(x)#

In order for #e^x# to be the only solution we need an extra condition such as #f(0) = 1#

Suppose #f(x)# is a well behaved function from #RR# to #RR# with #f(0) = 1#.

We can write the Maclaurin series for #f(x)# as:

#f(x) = sum_(n=0)^oo a_n x^n" "# for some constants #a_0, a_1,...#

Then:

#1 = f(0) = a_0#

and:

#f'(x) = sum_(n=0)^oo (d/(dx) a_n x^n)#

#color(white)(f'(x)) = sum_(n=0)^oo (n a_n x^(n-1))#

#color(white)(f'(x)) = sum_(n=0)^oo (n+1) a_(n+1) x^n#

Since we want #f'(x) = f(x)#, we have:

#sum_(n=0)^oo (n+1) a_(n+1) x^n = sum_(n=0)^oo a_n x^n#

So equating the coefficients, we find:

#{ (a_0 = 1), (a_(n+1) = a_n/(n+1)" for "n >= 0") :}#

So:

#1/a_0 = 1/1 = 0!#

#1/a_1 = 1/a_0 = 1!#

#1/a_2 = 2/a_1 = 2!#

#1/a_3 = 3/a_2 = 3!#

and:

#1/a_(n+1) = (n+1) (1/a_n) = (n+1)!#

Hence:

#f(x) = sum_(n=0)^oo x^n/(n!)#

which is one of the definitions of #e^x#

Then if #k# is any constant, we find:

#d/(dx) k e^x = k d/(dx) e^x = k e^x#

So:

#f(x) = k e^x#

is also a solution of #f(x) = f'(x)#

Footnote

There are some non well behaved functions that also satisfy #f'(x) = f(x)#

For example:

#f(x) = { ("undefined" " if " x=0), (abs(e^x)" if "x != 0) :}#

Answer:

#16 x^2 - 24 x y + 9 y^2 - 144 x + 8 y + 224 = 0#

Explanation:

The formula for the Distance from a Point to a Line is:

#d = |ax+by+c|/sqrt(a^2+b^2)" [1]"#

where #(x,y)# is the point and the line is of the form #ax+by+c = 0#.

The distance from any point #(x,y)# on the parabola to the line #3x+4y-1= 0# is:

#d = |3x+4y-1|/sqrt(3^2+4^2)#

#d = |3x+4y-1|/5" [2]"#

The distance from the focus #(3,0)# to any point, #(x,y)# on the parabola is:

#d = sqrt((x-3)^2+(y-0)^2)" [3]"#

The definition of a parabola is the locus of points equidistant from its directrix and its focus, therefore, we can derive the equation of the parabola by setting the right side of equation [2] equal to the right side of equation [3]:

#|3x+4y-1|/5 = sqrt((x-3)^2+(y-0)^2)#

Multiply both sides by 5:

#|3x+4y-1| = 5sqrt((x-3)^2+(y-0)^2)#

Square both sides:

#(3x+4y-1)^2 = 25((x-3)^2+(y-0)^2)#

Expand the squares:

#9x^2 + 24xy + 16y^2 - 6x - 8y + 1 = 25x^2 + 25y^2 - 150x + 225#

Simplify:

#-16 x^2 + 24 x y - 9 y^2 + 144 x - 8 y - 224 = 0#

Multiply both sides by -1:

#16 x^2 - 24 x y + 9 y^2 - 144 x + 8 y + 224 = 0#

The above is the standard Cartesian form for a conic section.

Here is a graph of the parabola, the focus and the directrix:

www.desmos.com/calculator

Answer:

#x^2/9-y^2/16=1#.

Explanation:

Recall that, for the Hyperbola # S : x^2/a^2-y^2/b^2=1#, the

asymptotes are given by, #y=+-b/a*x;" where, "a,b gt 0#.

In our Case, the asymptotes are, #y=+-4/3*x#.

#:. b/a=4/3...........................................................................(1)#.

Next, #(3sqrt2,4) in S rArr (3sqrt2)^2/a^2-4^2/b^2=1#.

#:. 18/a^2-16/b^2=1#.

Multiplying by #b^2#, we get, #18*b^2/a^2-16=b^2#.

By #(1)#, then, #18*(4/3)^2-16=b^2, i.e., b^2=16#.

#:. b=4 (because, b >0), and, (1) rArr a=3, or, a^2=9#.

With #a^2=9, and b^2=16#, the reqd. eqn. of the Hyperbola is,

#x^2/9-y^2/16=1#.

Answer:

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

Explanation:

Solve:
# x^2 − 4x ≥ 5 #

1) Examine the related equation to find the boundaries.
# x^2 − 4x ≥ 5 #
# x^2 − 4x - 5 = 0 #
# ( x - 5 ) ( x + 1 ) = 0 #
# ( x - 5 ) = 0 #
# x = 5 #
or
# ( x + 1 ) = 0 #
# x = -1 #

The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.

  • Since the inequality is inclusive ("non-strict") the boundary points are in the solution set.

A graph of the number line, with solid dots on (-1), and (5) will help in thinking about the problem.

To check for the solution, use test points, one from each region, in the inequality to see if they qualify:

Try (-2), (0), and (6). It is best not to pick numbers that make the arithmetic difficult.

In region, # -oo < x < -1 #
# ( -2 )^2 − 4( -2) ≥ 5 " ?"#
# 4 + 8 ≥ 5 " ?"#
# 12 ≥ 5 " YES"#

In region, # -1 < x < 5 #
# ( 0 )^2 − 4( 0 ) ≥ 5 " ?" #
# 0 ≥ 5 " NO" #

In region, # 5 < x < oo #
# ( 6 )^2 − 4( 6 ) ≥ 5 " ?" #
# 36 − 24 ≥ 5 " ?" #
# 12 ≥ 5 " YES" #

We must be careful in stating the solution inequalities. x can not be in both regions at once, since they are disconnected. There is a full region between the two regions in the solution, so we can't use "and", we must use "or".

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

2-D Graph of the inequality:
enter image source here

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