Video Transcript
In this video, we will learn how to
solve cubic equations using the cube root property. We begin by recalling that the cube
root of a number π written as shown is the number whose cube is π. In other words, the cube root of π
cubed is equal to π. We can use this to simplify or
evaluate expressions. For example, we know that two cubed
is equal to eight, which means that the cube root of eight is two.
This is not the only use of the
cube root, as we can also use this idea to solve equations. For example, imagine we are told
that the volume of a cube is 64 centimeters cubed. We can then say that the cube has
side lengths π₯ centimeters, giving us the equation π₯ cubed is equal to 64. Cube rooting both sides of this
equation, we know that the cube root of π₯ cubed is π₯. And the cube root of 64 is
four. We can therefore conclude that the
cube has side lengths of four centimeters.
At this stage, it is worth noting
that the equation π₯ cubed equals π will only have one solution for any real value
of π. This is because cubing a positive
number gives a positive answer, and cubing a negative number gives a negative
answer. This is different to solving the
equation π₯ squared equals π as this has two solutions when π is a positive real
number. For example, when π₯ squared is
equal to nine, we know that π₯ is equal to positive or negative three, since three
squared is equal to nine and negative three squared is equal to nine. Letβs now look at an example where
we need to solve a cubic equation.
Solve the cubic equation π₯ cubed
equals eight for all rational numbers.
We recall that a rational number is
a number that can be expressed as the quotient or fraction of two integers. In this question, we need to solve
the equation π₯ cubed equals eight and we will do so by firstly taking the cube root
of both sides. We know that the cube root of π₯
cubed is simply equal to π₯. And since two cubed is equal to
eight, the cube root of eight is equal to two. The solution to the cubic equation
π₯ cubed is equal to eight is therefore π₯ is equal to two.
In our next example, we will see
how to solve a cubic equation where the variable cubed is a fraction.
Solve π₯ cubed is equal to 27 over
eight.
In order to solve this equation, we
will begin by taking the cube root of both sides. We recall that the cube root of π₯
cubed is π₯. So π₯ is equal to the cube root of
27 over eight. Next, we recall that the cube root
of π over π, where π is nonzero is equal to the cube root of π over the cube
root of π. We know that since two cubed is
equal to eight, the cube root of eight is two. And since three cubed is 27, the
cube root of 27 is three. Since our equation simplifies to π₯
is equal to the cube root of 27 over the cube root of eight, then π₯ is equal to
three over two or three-halves. This is the single solution to the
equation π₯ cubed is equal to 27 over eight.
In our next example, we will solve
a cubic equation by first rearranging.
Given that π₯ exists in the set of
real numbers and negative π₯ over 10 is equal to 100 over π₯ squared, determine the
value of π₯.
In order to solve the given
equation, weβll begin by cross multiplying. This is the same as multiplying
both sides of the equation by 10π₯ squared. On the left-hand side, we have
negative π₯ multiplied by π₯ squared. And on the right-hand side, 100
multiplied by 10. This simplifies to negative π₯
cubed is equal to 1000. Multiplying through by negative one
so that our π₯-term is positive, we have π₯ cubed is equal to negative 1000. We can then take the cube root of
both sides of this equation. The cube root of π₯ cubed is
π₯. And noting that negative 10 cubed
is equal to negative 1000, then the cube root of negative 1000 is negative 10. If negative π₯ over 10 is equal to
100 over π₯ squared, then the value of π₯ is negative 10.
We can check this answer by
substituting our value of π₯ back in to the original equation.
So far in this video, we have only
doubt with simple equations involving cubes. However, it is possible for
operations to occur inside the cubic operation. In general, we can solve equations
of the form ππ₯ plus π all cubed plus π is equal to π provided that π is not
equal to zero and we can find the cube root of π minus π. We solve an equation of this type
using the following four steps.
Firstly, we subtract π from both
sides, giving us ππ₯ plus π all cubed is equal to π minus π. Secondly, we take the cube root of
both sides of the equation to get ππ₯ plus π is equal to the cube root of π minus
π. Next, we subtract π from both
sides such that ππ₯ is equal to the cube root of π minus π minus π. And finally, we divide through by
π so that π₯ is equal to the cube root of π minus π minus π all divided by
π. Letβs now consider how we can apply
this in practice.
Find the value of π¦ given that two
π¦ minus 14 all cubed minus 36 is equal to 28.
The equation in this question is
written in the form ππ₯ plus π all cubed plus π is equal to π. We know that we can solve equations
of this type by rearranging to make π₯ the subject. In this question, the variable is
π¦. So we will follow a similar method
in order to make π¦ the subject. We begin by adding 36 to both sides
of our equation. This gives us two π¦ minus 14 all
cubed is equal to 28 plus 36. The right-hand side simplifies to
64, and we can now take the cube root of both sides of this equation. On the left-hand side, we have two
π¦ minus 14. And since four cubed is equal to
64, the cube root of 64 is four. Our equation simplifies to two π¦
minus 14 equals four.
Next, we add 14 to both sides. Two π¦ is equal to four plus 14,
which is equal to 18. Finally, we can divide through by
two such that π¦ is equal to nine. If two π¦ minus 14 all cubed minus
36 is equal to 28, then the value of π¦ is nine. We can check this answer by
substituting π¦ is equal to nine back into our original equation. When we do this, we have two
multiplied by nine minus 14 inside our parentheses. This is equal to four, and four
cubed minus 36 is indeed equal to 28. This confirms that the solution to
the equation is π¦ is equal to nine.
In our final example, weβll solve a
similar problem. However, this time the coefficient
of the variable is negative.
Find the value of π₯ given that 15
minus three π₯ all cubed plus two is equal to 29.
In order to answer this question,
we will first rearrange the equation so that the cubed term is isolated on the
left-hand side. We do this by subtracting two from
both sides, giving us 15 minus three π₯ all cubed is equal to 29 minus two. 29 minus two is equal to 27. And we can now take the cube roots
of both sides of the equation. The cube root of 15 minus three π₯
all cubed is simply 15 minus three π₯. And since three cubed is 27, the
cube root of 27 is equal to three. Our equation simplifies to 15 minus
three π₯ is equal to three.
In order to solve for π₯, we can
now subtract 15 from both sides so that negative three π₯ is equal to three minus
15. This in turn simplifies to negative
three π₯ is equal to negative 12. And dividing through by three, we
have π₯ is equal to four. The value of π₯ that satisfies the
equation 15 minus three π₯ all cubed plus two is equal to 29 is four.
We will now finish this video by
recapping the key points. We saw in this video that we can
solve equations by taking the cube roots of both sides of the equation. In particular, if π₯ cubed is equal
to π, then π₯ is equal to the cube root of π. We also saw that unlike the square
root, taking cube roots of both sides of an equation gives a unique solution. Finally, to solve a cubic equation
of the form ππ₯ plus π all cubed plus π equals π, where π, π, π, and π are
constants and π is not equal to zero, we rearrange the equation to isolate π₯. This gives us π₯ is equal to the
cube root of π minus π minus π all divided by π.