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Question Video: Solving Quadratic Equations with Imaginary Roots Mathematics • First Year of Secondary School

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Solve the equation π‘₯Β² = βˆ’16.

02:17

Video Transcript

Solve the equation π‘₯ squared equals negative 16.

To solve an equation like this, we do begin by solving as we would any equation with real solutions by performing a series of inverse operations. In this case, we’re going to find the square root of both sides of the equation. Before we do though, we choose to rewrite negative 16 slightly. We’re going to write it as 16𝑖 squared. And we’ll see why we do this in a moment. But for now, it works because 𝑖 squared is equal to negative one. And this means that 16𝑖 squared is 16 times negative one which is negative 16.

And now that we have the equation π‘₯ squared equals 16𝑖 squared, we can now find the square root of both sides of the equation, remembering that we can take both the positive and negative root of 16𝑖 squared. The square root of π‘₯ squared is π‘₯. So π‘₯ is equal to the positive and negative square root of 16𝑖 squared. And during this next step, it’s going to become evident why we chose to write negative 16 as 16𝑖 squared. We can split the square root of 16𝑖 squared into the square root of 16 times the square root of 𝑖 squared.

The square root of 16 is four and the square root of 𝑖 squared is simply 𝑖 so in turn we can see that π‘₯ is equal to plus or minus four 𝑖. The solutions to the equation π‘₯ squared equals negative 16 are four 𝑖 and negative four 𝑖. And we should now be able to see why we did write negative 16 as 16𝑖 squared. It made these final steps a little easier to deal with.

And of course, we can check these solutions by substituting them back into the original equation. Let’s try this for π‘₯ equals four 𝑖 first. π‘₯ squared is four 𝑖 squared. And of course, that’s four 𝑖 times four 𝑖. Multiplication is commutative. It can be performed in any order. So we can rewrite this as four times four times 𝑖 times 𝑖. Four multiplied by four is 16 and 𝑖 times 𝑖 is 𝑖 squared. And since 𝑖 squared is negative one, π‘₯ squared is negative 16 as required.

We can repeat this process for π‘₯ equals negative four 𝑖. π‘₯ squared is negative four 𝑖 times negative four 𝑖, which can in turn be written as negative four times negative four times 𝑖 times 𝑖. Once again, since negative four times negative four is positive 16, we get 16𝑖 squared which is negative 16 as required.

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