WEBVTT
Kind: captions
Language: en

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&gt;&gt; Welcome to the Cypress College
Math Review on solving quadratic equations.

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There are four methods for
solving quadratic equations.

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Solve by taking square roots, which is
also called extracting square roots.

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Solve by factoring using the zero product rule.

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Solve by completing the square.

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And solve by using the quadratic formula.

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We are going to review each
of these different methods.

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Objective one, solving by taking square roots.

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Let's review this method.

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First of all, in this equation we
notice that the only X is in a square.

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This equation will solve very
easily using this method.

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You want to isolate the perfect square.

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The perfect square is the X squared.

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Notice that there are no
other Xs in the equation.

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So first we add 3 to both sides.

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Then we divide by 2.

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Now we have the perfect square
completely isolated.

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The next step is to take the
square root of both sides.

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Now on the left-hand side you get the absolute
value of X which will give you plus or minus X.

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And when you solve for X you get
plus or minus the square root of 5.

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No one actually shows those steps.

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Most people go straight from this
equation, X squared equals 5,

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to X equals plus or minus the square root of 5.

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For this problem the perfect
square is the square of X plus 4.

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Notice that there are no
other Xs in the equation.

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So our first step is to isolate that.

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So we add 5 to both sides.

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Then divide both sides by 2.

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Now we take the square root of both sides.

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So X plus 4 is equal to plus
or minus the square root of 12.

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We need to take out any perfect
squares that we can out of

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that radical, and 4 is a perfect square.

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So we have our answer of X equals negative 4
plus or minus 2 times the square root of 3.

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Pause the video and try these problems.

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Objective two.

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Solve by factoring.

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In this equation we not only have
the quadratic term 5X squared,

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but we also have the linear term negative 14X.

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So X is not just in that
perfect square X squared.

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So our first method won't work.

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To solve by factoring, you need
to move all terms to one side.

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I always like to have a positive
leading coefficient

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when I'm factoring so I make sure that happens.

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I get a positive 5X squared.

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And the right-hand side you have to have a zero.

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It has to equal zero.

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The next step is to factor your trinomial.

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To get 5X squared I would have 5X and
X. There's different options for 8.

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I could have 1 times 8 or 2 times 4.

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It will work out that I need
a 4 here and a 2 there.

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This is 4X and this is 10X.

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Those would add to get 14X.

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The signs need to be the same
because this is a positive.

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And they both need to be plus to be able
to make it so that I have positive 14X.

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Now I have it factored.

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By the zero product rule we know
that if A times B is equal to zero,

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then either A is equal to
zero or B is equal to zero.

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So I have the quantity 5X plus 4 times
the quantity X plus 2 is equal to 0.

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That means that either the first factor is
equal to 0 or the second factor is equal to 0.

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Now we solve each of these individual equations.

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So we have X equals negative
4/5 or X equals negative 2,

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giving us the two answers to this equation.

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Our next example we will
solve by factoring again.

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We want a positive leading coefficient
so we move everything to the left.

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Equals 0. Now we factor.

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So we have 5X and 3X and we
need a 2 here to give us 6X.

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And a 4 here to give us 20X.

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The signs are going to be
different because the 8 is negative.

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And I'm going to need a negative 20X and a
positive 6X to get the negative 14X that I need.

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Now the left-hand side is factored.

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Now we're going to use the zero product rule.

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Since A times B is equal to 0,
either the first factor is equal to 0

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or the second factor is equal to 0.

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Then we solve each of these equations.

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So we get X equals negative 2/5 or X equals 4/3
which are the two solutions to this equation.

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Pause the video and try these problems.

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Objective three.

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Solve by completing the square.

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To solve by completing the square, we first
move the constant term to the other side.

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So we add 3 to both sides for this equation.

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And then we divide by the leading
coefficient so that you have a 1X squared.

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In this equation we already have 1X
squared so we don't need to do that.

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Next you complete the square.

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To complete the square you take half
of the negative 10 which is negative 5,

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square it, and add it to both sides.

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So to complete the square we
have to add 25 to both sides.

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What you do to one side you
have to do to the other.

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You're always adding when
you complete the square.

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Next you write the left-hand
side as a perfect square.

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X squared.

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This would be an X. This
is minus so this is minus.

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And half of 10 was 5.

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Check it out.

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If you square X minus 5 you do
get X squared minus 10X plus 25.

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The right-hand side is 28.

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Now we use our first method.

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We take the square root of both sides.

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So X minus 5 is equal to plus
or minus the square root of 28.

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Now we simplify that radical
and that's 4 times 7.

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So the right-hand side becomes plus
or minus 2 times the square root of 7.

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We solve for X by adding 5 to both sides.

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So we get 5 plus or minus 2
times the square root of 7.

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Sometimes it's okay to write
your answers like that.

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Other times your instructor
wants them written separately.

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So 5 plus 2 times the square root of 7
or 5 minus 2 times the square root of 7.

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Those are the solutions to this equation.

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Let's solve this equation
by completing the square.

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So the first step is to move the
constant term to the other side.

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So we add 8 to both sides.

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The next step is to divide by the leading
coefficient so that we have a 1X squared.

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Be sure you divide every single term on both
sides of the equation by that coefficient.

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So we have X squared plus 6X is equal to 2.

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Now we complete the square.

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Half of 6 is 3, squared is 9.

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So we add 9 to both sides.

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The left-hand side becomes a perfect square.

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X squared becomes X plus, plus, half of 6 was 3.

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The right-hand side is 11.

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Now we take the square root of both sides.

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So X plus 3 equals plus or
minus the square root of 11,

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and the square root of 11 cannot be simplified.

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So our answers are negative 3 plus
or minus the square root of 11.

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Pause the video and try these problems.

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Objective four.

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Solve by using the quadratic formula.

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So for this equation we're going
to solve using that method.

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So we first move all terms to
one side and get them organized.

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So we add 12X to both sides.

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We have to have it equal to 0.

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So we must have it in the form AX
squared plus BX plus C equals 0.

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From that we can determine that
A is 3, B is 12, and C is 7.

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Next we write down the formula.

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So X equals negative B plus or minus the
square root of B squared minus 4AC all over 2A.

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Use parentheses when you're
replacing variables with numbers.

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So I have parentheses everywhere I see a letter.

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Now I substitute in.

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So B is 12, A is 3, and C is 7.

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Now we simplify this.

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So we have X equals negative 12 plus or minus
the square root of -- And 12 squared is 144.

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And then minus 84.

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Then that's all over 6.

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So we get negative 12 plus or minus
the square root of 60 all over 6.

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Be careful not to cancel the 12 and the 6.

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You can only cancel factors, and
the numerator is not factored.

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We have to simplify that radical.

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60 is 4 times 15.

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So we have 2 times the square root of 15.

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Now we can factor out a 2.

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We do so and we end up with negative 6 plus
or minus the square root of 15 all over 6.

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Now we can cancel.

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2 goes in to here once and
2 goes in to here 3 times.

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So our answers to this equation are negative 6
plus or minus the square root of 15 all over 3.

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We're going to solve this using
the quadratic formula again.

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So we move all terms to one side.

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Have it equal to 0.

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So we have it in the form AX
squared plus BX plus C equals 0.

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That tells us that A is 5, B is
negative 8, and C is negative 3.

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Next we write the formula.

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X equals negative B plus or minus the square
root of B squared minus 4AC all over 2A.

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Then we use parentheses since we're
replacing variables with numbers.

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So we have that B is negative
8, A is 5, and C is negative 3.

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Now we simplify this.

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So we have positive 8 plus or
minus the square root of 64

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because that's negative 8 quantity
squared, and then watch out for your signs.

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We've got two negatives so we
end up with plus 60 all over 10.

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So we have X equals 8 plus or minus
the square root of 124 all over 10.

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124 is 2 times 62 and 62 is 2 times 31.

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So X equals 8 plus or minus the square root of 4

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because we have 2 squared
times the 31 all over the 10.

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I'm trying to simplify the radical.

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So we get 8 plus or minus 2 rad 31 all over 10.

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Now we factor out a 2 and we're left with 4
plus or minus the square root of 31 all over 10.

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Now you can cancel.

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So many people coming back over here
will cancel with the 8 and the 10.

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And you cannot cancel the 8 with the 10
because the 8 is not a factor of the numerator.

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Looking over here on the right,
the 2 is a factor of the numerator.

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Since 2 is a factor of the numerator,
I can cancel and I get 1 over 5.

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So the solution to this equation is 4 plus
or minus the square root of 31 all over 5.

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Pause the video and try these problems.

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Objective five.

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Determine the best method to
solve a quadratic equation.

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In the equation AX squared plus BX plus C
equals 0, AX squared is a quadratic term,

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BX is called the linear term,
and C is the constant.

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If your equation does not have a
constant term, then solve by factoring.

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That's definitely going to
be the easiest method.

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In this equation we move all terms to one side
and we notice that there's no constant term.

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Every single term has an X in it.

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So we factor, set each factor
equal 0, and then solve.

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Any time the equation factors easily
then that's the easiest method to solve.

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Solve by factoring if the
equation factors easily.

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If your equation does not have a linear
term, then solve by taking square roots.

00:17:13.016 --> 00:17:17.456
The only place you see the X is
in the perfect square X squared.

00:17:18.776 --> 00:17:21.246
So you're going to solve by taking square roots.

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So in this equation we move 9
to the other side, divide by 4,

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and take the square root of both sides.

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So our answers are X equals plus or minus 3/2.

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If your equation looks like 1X squared
plus BX plus C equals 0, and B is even,

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then completing the square is really fast.

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Any time you have a 1X squared and you
have an even linear coefficient, B is even,

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then you should complete the square.

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So for this equation we have 1X squared
minus 8X, and then we have a constant term.

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The constant term can be on either side,
but definitely for this equation solve

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by completing the square is the easiest.

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So we have the constant on the other side.

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That's the first step, already done for us.

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Now we complete the square.

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Half of 8 is 4, squared is 16.

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We add 16 to both sides.

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The left-hand side we write as a perfect square.

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Then we take the square root
of both sides of the equation.

00:18:41.236 --> 00:18:45.906
So we have X minus 4 equals plus
or minus the square root of 19.

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We add 4 to both sides and we get our solutions,
4 plus or minus the square root of 19.

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If your equation looks like this where the only
place you see the X is in a perfect square,

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then solving by taking square
roots is the easiest method.

00:19:10.076 --> 00:19:13.566
So we want to get the perfect
square by itself by adding 4

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to both sides and then dividing by 3.

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Then we take the square root of both sides.

00:19:20.746 --> 00:19:27.976
So we get X plus 2 equals plus or minus the
square root of 4, but the square root of 4 is 2

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so now we have to write it as two separate
equations because our radical disappeared.

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We subtract 2 from both sides and we get
X equals negative 2 plus 2 which is 0

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or X equals negative 2 minus
2 which is negative 4,

00:19:45.386 --> 00:19:48.336
and those are the two solutions
to this equation.

00:19:51.236 --> 00:19:58.466
If your equation involves fractions,
then multiply both sides of the equation

00:19:58.466 --> 00:20:02.056
by the appropriate quantity, which of
course is the least common denominator,

00:20:03.116 --> 00:20:04.766
to get rid of the fractions.

00:20:05.666 --> 00:20:09.836
Then use whatever method is easiest
to solve the resulting equation.

00:20:11.056 --> 00:20:13.416
If your equation involves decimals,

00:20:14.106 --> 00:20:17.166
then the quadratic formula is
probably the easiest method.

00:20:21.796 --> 00:20:26.046
If you're working on a word problem
or any problem where you're allowed

00:20:26.086 --> 00:20:31.606
to use an approximation, then the
quadratic formula is usually the fastest.

00:20:32.236 --> 00:20:35.096
Substitute A, B, and C in to the formula,

00:20:35.496 --> 00:20:38.656
then pull out a calculator
and approximate the answers.

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No need to simplify.

00:20:43.420 --> 00:20:45.280
Pause the video and try these problems.

00:20:49.140 --> 00:20:50.620
Objective six.

00:20:51.146 --> 00:20:52.836
Practice solving by any method.

00:20:54.196 --> 00:20:58.366
For each of the problems in the
exercise set, choose the easiest method,

00:20:59.540 --> 00:21:04.080
then solve the quadratic equations.

00:21:05.320 --> 00:21:07.860
Pause the video and try these problems.

