### Video Transcript

Find the solution set of the equation two π₯ squared minus 20 equals zero in the real numbers, giving values to one decimal place.

We could find the solution set a few different ways. One way would be to isolate π₯ squared and then take the square root. So we can do so by adding 20 to both sides of the equation first. Now, we have two π₯ squared equals 20. So we need to divide both sides of the equation by two. So π₯ squared is equal to 10.

Now to solve for π₯, we need to square root both sides of the equation. And we find that π₯ is equal to positive and negative 3.16. But we need to round to one decimal place. So we will either keep the one a one or round it up to a two. So we look to the number to the right. And since six is five or larger, we will round the one up to a two. So we have that π₯ is equal to plus or minus 3.2. And 3.2 and negative 3.2 are both in the real numbers. So these would be our final answers.

Letβs go ahead and solve this another way. We could have factored by taking out a two as our GCF. And now, thereβs nothing left to factor. So we can set each of our factors equal to zero. If we set two equal to zero, thatβs not even a true statement. Two is not equal to zero. So this does not give us a solution. There would need to be a variable with it. And then we set π₯ squared minus 10 equal to zero. So we would need to add 10 to both sides of the equation. So π₯ squared would be equal to 10. And then square rooting both sides, we get that π₯ is equal to plus or minus 3.16. And after rounding, we would get the 3.2.

Lastly, we also could have used the quadratic formula, where π₯ is equal to negative π plus or minus the square root of π squared minus four times π times π all divided by two π. π, π, and π are found here: ππ₯ squared plus ππ₯ plus π. So for our equation, we have two π₯ squared minus 20 equals zero. π will be negative 20. Itβs the constant. And then, π would need to be two. And since thereβs no π, π must be zero. So once again, π is equal to two, π is equal to zero, and π is equal to negative 20.

So letβs go ahead and plug these in. So now that weβve plugged in the values for π, π, and π, we need to simplify. We have zero plus or minus the square root of 160 all divided by four. And the square root of 160 is 12.6491106407. And the zero in front really isnβt doing anything. So we can erase it. So we need to take our number and divide by four. And we find that π₯ is equal to plus or minus 3.162277. And after rounding one decimal place, we would still get 3.2. So once again, our final answer will be 3.2 and negative 3.2.