Question 162565
<pre><font size = 4 color = "indigo"><b>
Solve the simultaneous equations: 

{{{system(32x^2+15y^2=2112,7x^2-3y^2=60)}}}

Multiply the second equation through by 5 to
make the {{{y^2}}}-terms cancel when we add
the equations vertically term by term:

{{{system(32x^2+15y^2=2112,35x^2-15y^2=300)}}}

Adding them term by term:

{{{67x^2=2412}}}
Divide both sides by 67
{{{x^2=2412/67}}}
{{{x^2=36}}}
{{{x^2-36=0}}}
Factor:
{{{(x-6)(x+6)=0}}}

{{{matrix(2,3, x-6=0, ",", x+6=0,  x=6, ",", x=-6)}}}

Substitute {{{x=6}}} into 

{{{7x^2-3y^2=60}}}
{{{7(6)^2-3y^2=60}}}
{{{7(36)-3y^2=60}}}
{{{252-3y^2=60}}}
{{{-3y^2=-192}}}
Divide both sides by {{{-3}}}:

{{{y^2=64}}}
{{{y^2-64=0}}}
{{{(y-8)(y+8)}}}

{{{matrix(2,3, y-8=0, ",", y+8=0,  y=8, ",", y=-8)}}}

So we have these solutions for {{{x=6}}},

(x,y) = (6,8)
(x,y) = (6,-8)

Substitute {{{x=-6}}} into 

{{{7x^2-3y^2=60}}}
{{{7(-6)^2-3y^2=60}}}
{{{7(36)-3y^2=60}}}
{{{252-3y^2=60}}}
{{{-3y^2=-192}}}
Divide both sides by {{{-3}}}:

{{{y^2=64}}}
{{{y^2-64=0}}}
{{{(y-8)(y+8)=0}}}

{{{matrix(2,3, y-8=0, ",", y+8=0,  y=8, ",", y=-8)}}}

So we have these solutions for {{{x=-6}}},

(x,y) = (-6,8)
(x,y) = (-6,-8)

So there are four solutions:

(x,y) = (6,8)
(x,y) = (6,-8)
(x,y) = (-6,8)
(x,y) = (-6,-8)

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What is the substitution for the quadratic formula for equation:
                                   

{{{8x^2+3a^2=10ax}}}

{{{8x^2-10ax+3a^2=0}}}

Since this already contains a small letter "a", we'll
write the quadratic formula using all CAPITAL LETTERS:

{{{x = (-B +- sqrt( B^2-4*A*C ))/(2*A) }}} 

where {{{A=8}}}, {{{B=-10a}}}, and {{{C=3a^2}}}

{{{x = (-(-10a) +- sqrt( (-10a)^2-4*(8)*(3a^2) ))/(2*(8)) }}}

{{{x = (10a +- sqrt( 100a^2-96a^2) )/16 }}}

{{{x = (10a +- sqrt(4a^2) )/16 }}}

{{{x = (10a +- 2a)/16 }}}

{{{x=2a(5 +- 2)/16}}}

{{{x=a(5+-2)/8}}}

Using the +,

{{{x=a(5+2)/8=7a/8}}}

Using the +,

{{{x=a(5-2)/8=3a/8}}}

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What would the problem like at the beginning:  

{{{system(x^2+y^2=65, (1/2)xy=14)}}}

To clear of fractions, multiply second equation through
by 2:

{{{system(x^2+y^2=65, xy=28)}}}

Solve the second equation for y:

{{{xy=28}}}
{{{y=28/x}}}

Substitute that in the first equation:

{{{x^2+y^2=65}}}
{{{x^2+(28/x)^2=65}}}
{{{x^2+784/x^2=65}}}
{{{x^4+784=65x^2}}}
{{{x^4-65x^2+784=0}}}
{{{(x^2-49)(x^2-16)=0}}}
{{{(x-7)(x+7)(x-4)(x+4)=0}}}

{{{matrix(2,7,  x-7=0, ",", x+7=0, ",", x-4=0, ",", x+4=0,  x=7, ",", x=-7, ",", x=4, ",", x=-4)}}}  

Substitute each of those into:

{{{xy=28}}}

{{{matrix(4,7,  x=7, ",", x=-7, ",", x=4, ",", x=-4, 
              (7)y=28, ",", (-7)y=28, ",", (4)y=28, ",", (-4)y=28,
                7y=28, ",", -7y=28, ",",4y=28, ",",-4y=28, 
                 y=4, ",",   y=-4, ",",    y=7, ",",    y=-7) 


)}}}

So there are 4 solutions:

(x,y)=(7,4)
(x,y)=(-7,-4)
(x,y)=(4,7)
(x,y)=(-4,-7)

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if the discriminant of a complete quadratic equation is 8, 
what is the nature of its roots.

When the discriminant is positive, there are two different real solutions.
When it is 0 there is exactly one real solution
When it is negative there are no real solutions.

8 is positive so there are two different real solutions.

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separate 72 into two parts so that the first part is the square of the second. let x represent the first part, what is the equation that would solve the problem.

The other part of {{{72}}} is {{{72-x}}}

First part: {{{x}}}
Second part: {{{72-x}}}

>>...the first part is the square of the second part...<<

So  

{{{x = (72-x)^2}}}
{{{x = (72-x)(72-x)}}}
{{{x = 5184-72x-72x+x^2}}}
{{{x = 5184-144x+x^2}}}
{{{0 = 5184-145x+x^2}}}
{{{0 = x^2-145x+5184}}}
{{{0 = (x-81)(x-64)}}}

{{{matrix(2,3, x-81=0, ",", x-64=0,  x=81, ",", x=64)}}}

So if x = first part = 81, then second part = 72-81=-9

That's kind of weird, separating 72 into 81 and -9, but
they do sum to 72, and certainly 81 is the square of -9.

Now if x = first part = 64, then second part = 72-64=8

That's not weird, separating 72 into 64 and 8, and
they do sum to 72, and certainly 64 is the square of 8.

So there are two answers, a weird one and one not weird at all.

Edwin</pre>