 
In the PreTrig review part 11 we go over techniques for solving equations
 
including the quadratic formula
 
and the zero product property
 
let's look at example a we have a rational equation
 
2 over 5x,  equals 4. all we need to do here is cross multiply
so the 5x is going to multiply the 4 so we get 2 is equal to
20 x
we're going to divide both sides by 20
and we will get
that x is equal to
one 10th
moving on to example b
 
we have a quadratic
x squared is equal to 20 plus 8x so the first step is to get everything on one side so 
we'll leave the x squared where it is 'cause it's positive
we're going to subtract the 8x from both sides so this will become -8x on the left hand side
and we have to subtract 20
and then the right hand side is now
zero so we're looking to factor
we need two numbers that will multiply to be -20
but will add to be -8
so we start going through the factors of 20
two we would have one and -20
two and -10 and that's it so we have a positive two and a -10 that we want to use
and since the leading coefficient is a one
we know that we can just write these
as our factors so we have x+2 
times x-10 
equal zero
now we're going to make use of the zero product property
and we set each factor equal to zero
so x+2 has to equal zero
or x-10 has to equal zero
and if we solve each one the first one we get
x is equal to -2 solving the second one we get
x is equal to 10
moving onto an example c
 
again we have a quadratic
so let's take a look at the factoring and we would need
two numbers that multiply
to be -6
and that would add to be
-6 so as we go through
we've got one and -6, 2 and -3
neither one of these adds up to be -6
so that means that the quadratic is prime
when the quadratic is prime
we're going to have to use the quadratic formula to solve it
so now we know we've got
x
is going to be negative
-6
plus or minus the square root
of -6 squared
-4
times two
times -3
divided by
two times two
well let's clean this up
we've got six plus or minus the square root of
36 and then there's a minus sign here and a -3 here so that's going to turn into plus
four times two is eight times three
24.  36 plus 24 is equal to
60 so we've got the square root of 60
and in the denominator we have a four
remember our list of perfect squares we went over in an earlier part of the review
and we know that if four goes into 60
so if we come off on the side here radical 60
will be radical four
times radical 15 which is two rad 15
so we can rewrite
this as six plus or minus
two radical 15
divided by four
now six and two have a common factor
which is 2.  So we can factor out a two and we would have three
plus or minus radical 15
divided by four
and now we can cancel we have
two goes into two once, two goes into four twice
and finally we have
that x
is equal to three
plus or minus radical 15
divided by two
 
in example d we have an equation that involves a radical so the first step is to isolate the radical
 
so we can have have x -3
is equal to radical
x plus three
it's always a good idea to make sure that your radical is positive, it's one less negative sign to have to keep track of
now we'll use the technique of squaring both sides
so we're going to take x-3 squared
is equal to radical x+3
squared and of course remember to be careful when you square both sides you know 
you're going to have to check for extraneous solutions
on the left hand side
we get x squared - 6x plus nine
and on the right hand side we get x+3
we're going to simplify our equation by combining like terms
so we're going to bring the x to combine with the -6x and the three to combine with the nine so now we have x squared -7x
plus six is equal to zero
and this is a pretty easy factorization
(x -1)(x-6)
is equal to zero
now we're going to use the zero product property
and we have x -1 is equal to zero
or x -6 is equal to zero
which means we get x is equal to one
or x is equal to six
now remember because we squared both sides we have to check for extraneous solutions
so we plug in always to the original equation and let's check
x is equal to one
and if we plug that into the original equation
we get one minus
the square root of one plus three and we're asking ourselves if that equals
three
so we have one minus
well this is gonna be the square root of four which of course is two
and 1 - 2
is -1 which definitely does not equal three
therefore we have to reject the x equals one solution
now let's go ahead and check
x equals six
into the original equation
so we have six minus the square root
of six plus three
and we're asking ourselves if that is equal to three
so we have six minus the square root of nine
and the square root of nine
is three and yes that does equal three so this equation has only one solution
and that solution is six
