Question Video: Solving Cubic Equations over the Set of Rational Numbers | Nagwa Question Video: Solving Cubic Equations over the Set of Rational Numbers | Nagwa

Question Video: Solving Cubic Equations over the Set of Rational Numbers Mathematics • Second Year of Preparatory School

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If possible, find the value of π‘₯ which satisfies βˆ’0.216π‘₯Β³ = 27, given that π‘₯ is in β„š.

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Video Transcript

If possible, find the value of π‘₯ which satisfies negative 0.216π‘₯ to the third power equals 27, given that π‘₯ is in the set of the rational numbers.

To solve this equation, we’re going to start by rearranging. Since we have negative 0.216 multiplied by π‘₯ to the third power, then we do the inverse operation which is dividing by negative 0.216. At this point, if we had a calculator, it then becomes very simple to put in this value, 27 divided by negative 0.216, and then take the cube root. However, let’s see if we can use a non-calculator method and see if we can write these values of values with exponents of three.

We can start by writing our negative 0.216 as a fraction. This will be negative 216 over 1000. We now have a division involving fractions. We recall that we change our division to multiplication, and we flip the numerator and denominator. The negative can be included on either the numerator or denominator, but we typically include it as part of the numerator.

Now, let’s see if we can write our values of 27, negative 1000, and 216 as values that have exponents of three. 27 can be written as three to the power of three or three cubed since three times three times three is 27. Negative 1000 can be written as negative 10 all cubed. And 216 can be written as six cubed. We have now got every term on the right-hand side as a power of three. We can therefore use the exponent rule, π‘₯ to the power of π‘Ž 𝑦 to the power of π‘Ž equals π‘₯𝑦 all to the power of π‘Ž.

This means that we can take our expression on the right-hand side, where every exponent or power is π‘Ž equal to three, and write it as a single product to the power of three. So, π‘₯ cubed is equal to three times negative 10 over six to the power of three. To find the value of π‘₯ then, we take the cube root of both sides of our equation.

And then, since the cube root of a value cubed is simply the value itself, we have π‘₯ equals three times negative 10 over six. Simplifying this by taking three as a common factor, we have one times negative 10 over two or simply negative 10 over two. So, π‘₯ is equal to negative five. As negative five is indeed a rational number, then we have found π‘₯ equals negative five, the value which satisfies the equation, and we did so using a non-calculator method.

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