Question 202134
the quadratic formula is:
x = {{{(-b +- sqrt(b^2-4*a*c))/(2*a)}}}
-----
the general form of the quadratic equation is:
{{{a*x^2 + b*x + c = 0}}}
-----
to use the quadratic equation, you find the values for a, b, and c, and plug them into the quadratic equation and solve.
-----
your first problem is:
{{{x^2 - 3x = 4x -1}}}
first you need to convert it to the standard form.
subtract 4x from both sides of the equation and add 1 to both sides of the equation to get:
{{{x^2 - 7x + 1 = 0}}}
a = 1
b = -7
c = 1
plug these values into the quadratic equation to get:
x = {{{ (-(-7) +- sqrt ((-7)^2 - 4*1*1))/(2*1)}}}
which becomes:
x = {{{ (7 +- sqrt(45))/2) }}}
this answer works as is but you can reduce it further by simplifying the number under the square root sign as follows:
x = {{{ ( 7 +- sqrt(3*3*5))/2 }}}
which reduces to:
x = {{{ ( 7 +- 3*sqrt(5))/2 }}}
all you need to do now is verify the answer is correct by plugging those values for x into the original equation to see if the equation is true.  I cheated by using my calculator, but the answers are verified to be true so these answers are good.
-----
your second problem is:
x^2 -5x - 1 = -7 
-----
you do the same thing.
convert it to standard form.
get the values of a, b, and c.
plug them into the formulas and solve.
-----
i'll solve without explanation this time and you can follow along or try to solve it on your own.
x^2 - 5x - 1 = -7
add 7 to both sides to get:
x^2 - 5x + 6 = 0
-----
a = 1
b = -5
c = 6
x = {{{(-b +- sqrt(b^2-4*a*c))/(2*a)}}}
x = {{{(-(-5) +- sqrt((-5)^2-4*1*6))/(2*1)}}}
which becomes
x = {{{(5 +- sqrt(25-24))/2}}}
which becomes
x = {{{(5 +- sqrt(1))/2}}}
which becomes
x = {{{(5 +- 1)/2}}}
which becomes
x = 3 or x = 2
-----
plug these values into the original equations to see if the equations are true.
original equation is:
x^2 -5x - 1 = -7 
when x = 2 this becomes
4 - 10 - 1 = -7
which becomes
-6-1 = -7 which is true.
when x = 3 this becomes
9 - 15 - 1 = -7
which becomes
-6 -1 = -7 which is also true.
the answers are good.
x = {{{(5 +- 1)/2}}}
-----
your last equation needs to be solved by completing the squares.
this is a different technique but gets you to the same answer.  
you could also use the quadratic equation here as well, but sometimes completing the square is easier.
-----
your last equation is:
{{{x^2 + 8x + 2 = 0 }}}
you would subtract 2 from both sides of the equation to get:
{{{x^2 + 8x = -2}}}
you take 8/2 = 4 and use that value to complete the square.
you take {{{(x+4)^2}}} to get:
{{{x^2 + 8*x + 16}}}.
so what you have is:
{{{(x+4)^2 = x^2 + 8*x + 16}}}.
if you subtract 16 from that, you have:
{{{x^2 + 8*x =(x+4)^2 - 16}}}
you can now go back to your original equation after modification which was:
{{{x^2 + 8*x = -2}}}
and substitute {{{(x+4)^2 - 16}}} for {{{x^2 + 8*x}}} to get:
{{{(x+4)^2 - 16 = -2}}}
you add 16 to both sides of the equation to get:
{{{(x+4)^2 = 14}}}
you have just completed the square.
solving by taking the square root of both sides and you will get:
{{{(x + 4)}}} = +/- {{{sqrt(14)}}}
which becomes:
{{{ x = -4 +- sqrt(14) }}}
you verify that this answer is valid by plugging it into the original equation which was:
{{{x^2 + 8*x + 2 = 0}}}
the values check out ok so the answer is good.
note that if you had solved this equation using the quadratic formula, you would have gotten the same answer.
-----
note that the standard form of the quadratic equation is:
{{{ax^2 + bx + c}}}.
to complete the square, you need to move the c over to the right side and you need to divide both sides of the equation by a, if a is not already 1.
-----
you will wind up with:
{{{ax^2 + bx = -c}}}
before dividing by a if it is not 1 to start with.
if a is not one, you will wind up with:
{{{(ax^2 + bx)/a = -c/a}}}
which is the same as:
{{{x^2 + (bx)/a = -c/a}}}
-----
i'll show you how that works with a simple example.
once you have the standard form for completing the square, the operations are the same as above.
-----
take {{{3x^2 + 9x + 6 = 0}}}
this is the standard form of the quadratic equation of {{{ax^2 + bx + c}}}.
note that a is not = to 1
first step:
move the c over to the right hand side by subtracting it from both sides of the equation.
you get:
{{{3x^2 + 9x = -6}}}
-----
since the a coefficient is not 1 (it's 3), you need to divide both sides of the equation by 3 to make it 1.
it becomes:
{{{(3x^2 + 9x)/3 = -6/3}}} 
which becomes:
{{{x^2 + 3x = -2}}}
-----
now you're in the standard form for completing the square.
-----
take 1/2 the b factor which means take 1/2 of 3.
you get 3/2
your factor that will be squared is {{{(x + 3/2)}}}.
-----
you get:
{{{(x + 3/2)^2 = x^2 + 3x + (9/4)}}}
your left side of the equation was {{{(x^2 + 3x)}}}
this equals {{{(x + 3/2)^2 - (9/4)}}}
your equation which was:
{{{x^2 + 3x = -2}}} becomes:
{{{(x + 3/2)^2 - (9/4) = -2}}}
you add (9/4) to both sides to get:
{{{(x + (3/2))^2 = -2 + (9/4)}}}
this becomes:
{{{(x + (3/2))^2 = (-8/4) + (9/4)}}}
which becomes:
{{{(x + (3/2))^2 = (1/4)}}}
take the square root of both sides to get:
{{{x + (3/2)}}} = +/- {{{sqrt(1/4)}}}
this means:
{{{x = -(3/2) + sqrt (1/4)}}}
or
{{{x = -(3/2) - sqrt (1/4)}}}
or
x = both.
you wind up with:
x = -1 or x = -2 or both.
both values of x checked out in the original equation so they're both good.
-----