Video: Pack 2 β€’ Paper 1 β€’ Question 15

Pack 2 β€’ Paper 1 β€’ Question 15

02:37

Video Transcript

Make 𝑑 the subject of the formula 𝑠 equals the cube root of five over 𝑑 squared plus four.

So the key thing about this question is that 𝑑 is what we’re trying to make the subject of the formula. Okay, so we’ve got 𝑠 is equal to the cube root of five over 𝑑 squared plus four. So therefore, the first step is to actually subtract four from each side. So we get 𝑠 minus four is equal to the cube root of five over 𝑑 squared.

Then, next, what we actually want to do is the inverse of the cube root because we got the cube root of five over 𝑑 squared. And to actually enable us to do that, what we’re gonna do is actually cube both sides of our equation, which is gonna give us 𝑠 minus four all cubed is equal to five over 𝑑 squared.

And then, what we’re gonna do is actually multiply each side of the equation by 𝑑 squared cause we actually got that down as the denominator. We don’t want it as a denominator. So we’re gonna multiply each side by 𝑑 squared. So we get 𝑑 squared multiplied by 𝑠 minus four all cubed equals five.

Then, at this stage, we remind ourselves that it’s 𝑑 that we want as the subject to the formula. So what we’re actually gonna do is divide through by 𝑠 minus four all cubed. So we’re gonna have 𝑑 squared is equal to five over 𝑠 minus four all cubed. So now, as we’re looking for single 𝑑 as a subject to the formula, what we’re gonna do is actually square root each side because it’s the inverse of squared because we have 𝑑 squared.

So now, for this next step, what we’re actually gonna do is use a little rule we know. And that’s the root of π‘Ž over 𝑏 is actually equal to the root of π‘Ž divided by the root of 𝑏. So using this rule, we’re gonna get 𝑑 is equal to root five over root and then 𝑠 minus four all cubed.

So now, what we’re gonna do is actually use one of our index rules to actually simplify this further because we know that the π‘Žth root of π‘₯ is equal to one over π‘Ž. So therefore, we can actually rewrite this as 𝑑 is equal to root five over and then we’ve got 𝑠 minus four to the power of three then this all to the power of a half.

And then, we can actually use one more index law to just help us get to our final solution. And that index law is that π‘₯ to the power of π‘Ž to the power of 𝑏 is equal to π‘₯ to the power of π‘Žπ‘. So we actually multiply the powers. So therefore, we can say that if we make 𝑑 the subject to the formula 𝑠 equals the cube root of five over 𝑑 squared plus four, we get 𝑑 equals root five over 𝑠 minus four to the power of three over two.

And we got that last bit because we had 𝑠 minus four to the power of three to the power of a half. Then, we actually used the index law, multiplied them together. And three multiplied by a half gives us three over two.

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