Question Video: Solving Quadratic Equations with Complex Roots Mathematics

Find the solution set of βˆ’6π‘₯Β² + 5π‘₯ βˆ’ 5 = 0 over β„‚.

02:41

Video Transcript

Find the solution set of negative six π‘₯ squared plus five π‘₯ minus five equals zero over the set of complex numbers.

So what I’ve got here is a quadratic equation. And we want to find the solution set. And we know that, in fact, they’re gonna involve complex numbers. This isn’t gonna change how we’re going to solve it. So if we take a look at this quadratic, we can see that what we can use to solve it is the quadratic equation. And what the quadratic formula tells us is that π‘₯ is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ over two π‘Ž. And this is when we’ve got our quadratic in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero.

So the first thing before we can use our quadratic equation is to identify what our π‘Ž, 𝑏, and 𝑐 are in our quadratic. Well, our π‘Ž is negative six, our 𝑏 is five, and our 𝑐 is negative five, remembering that we always need to take into account the signs of the values we’re looking at. So when we substitute in our values, we’re gonna get π‘₯ is equal to negative five plus or minus the square root of five squared minus four multiplied by negative six multiplied by negative five all over two multiplied by negative six.

Well, what this is gonna give us is π‘₯ is equal to negative five plus or minus the square root of negative 95 over negative 12. And in fact, it’s usually at this point that we’d stop. And that’s because if we take a look inside our square root, we can see that the discriminant, because the discriminant is 𝑏 squared minus four π‘Žπ‘, is negative. And because we’ve got a negative value and then a square root sign, we know that there is not a real solution to the square root of a negative number.

However, in the question, it asks us to find the solution set over the set of complex numbers. So therefore, we can use that to help us find a solution. And that’s because if we use our rules of surds or radicals, we can rewrite root negative 95 as root 95 multiplied by root negative one. And we can recall that root negative one is equal to the imaginary number 𝑖. So therefore, root negative 95 can be rewritten as root 95𝑖. So therefore, we can rewrite this as π‘₯ is equal to five plus or minus root 95𝑖 over 12. And then we can split this up further to write out the solution set of negative six π‘₯ squared plus five π‘₯ minus five equals zero over the set of complex numbers. And that is five over 12 minus root 95 over 12𝑖 and five over 12 plus root 95 over 12𝑖.

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