Find the solution set of the equation 𝑥² − 8𝑥 − 2 = 9𝑥 + 8, giving values correct to three decimal places.

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### Video Transcript

Find the solution set of the equation 𝑥 squared minus eight 𝑥 minus two equals nine 𝑥 plus eight, giving values correct to three decimal places.

Before we can even begin to solve an equation like this, we want to rearrange it and put it in standard form. Standard form is 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. We can do that by subtracting nine 𝑥 from both sides of the equation. Eight 𝑥 minus nine 𝑥 equals negative 17𝑥. We’ll also subtract eight from both sides of the equation. Negative two minus eight equals negative 10. And we’ve taken everything away on the right side of the equation so that it’s equal to zero. This equation is now in standard form. 𝑎 equals one. The coefficient of 𝑥 squared is one. 𝑏 equals negative 17. And 𝑐 equals negative 10. We’re not going to be able to factor this equation. And I wanna show you how I know that factoring is not an option here.

When we factor, we look for two values that multiply together to equal negative 10 and when added together equals negative 17. Now these values might exist, but they’re going to be fractions and very difficult to find. We can’t solve them by using factors of 10, like one and 10 or two and five. Since we can’t solve by factoring, another strategy is to use the quadratic formula. When we have an equation in standard form, we can use the quadratic formula to solve and find the solution set. The quadratic formula looks like this, negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

Now we just plug in what we know for 𝑎, 𝑏, and 𝑐. Negative of negative 17 plus or minus the square root of negative 17 squared minus four times one, because 𝑎 equals one, times negative 10 — 𝑐 equals negative 10 — all over two times one. We can simplify this a little bit. The negative of negative 17 equals positive 17, plus or minus the square root, negative 17 squared equals 289. And then we have negative four times negative 10 times one, which ends up to be positive 40. All over two times one, which equals two. We can add 289 and 40 together to get 329. And this is the simplified form, 17 plus or minus the square root of 329 over two.

And from here, we would need a calculator to solve. And we would need to break it up into two equations. We need to know 17 plus the square root of 329 divided by two. And we also need to know 17 minus the square root of 329 divided by two. The first result from my calculator yields 17.56917. And the second result yields negative 0.56917 continuing.

We want to round these values to three decimal places. We look at the third decimal place. In both cases, there’s a nine. And the digit to the right, the fourth decimal place, is our deciding digit. It will tell us how to round. Because in both of these cases the deciding digit is less than five, we will round down. 𝑥 equals 17.569 and 𝑥 equals negative 0.569. When we write it as a solution set, we have brackets 17.569 comma negative 0.569 and then end brackets.

I want to show you one other way to solve this problem. And this will work if you can’t remember the quadratic formula. For this form, we’ll again need standard form, 𝑥 squared minus 17𝑥 minus 10. And then we’re going to complete the square. We move that 𝑐 value to the other side of the equation by adding it to both sides. We then have 𝑥 squared minus 17𝑥 equals 10. And I want to add 𝑏 over two squared to both sides of the equation. This middle value is 𝑏. Negative 17 divided by two squared is what we want to add to both sides. Negative 17 over two squared is added to both sides. On the left, we now have a square. We have 𝑥 minus 17 over two squared is equal to 10 plus negative 17 over two squared.

Let’s go ahead and distribute this square value, which means we have 10 plus negative 17 squared divided by two squared, divided by four. Now, if we wanted to add 10 to this negative 17 squared over four, we would need to write 10 as a fraction over four. 10 equals 40 divided by four. We now have 40 plus negative 17 squared. And negative 17 squared equals 289. And all of that is being divided by four. Bring down the left side of the equation. And, hopefully, by now you’re starting to see this really cool pattern emerge. 40 plus 289 is what’s under the radical in the quadratic formula. On one side, we’re dividing by four. And the other side we’re dividing by two. So let’s keep an eye on that.

To get 𝑥 by itself, we’ll need to take the square root of both sides of the equation. We can rewrite the square root of 40 plus 289 over four as the square root of 40 plus 289 over the square root of four. 40 plus 289 equals 329. And the square root of four equals two. At this point, we add 17 over two to both sides of the equation. We need to know that the square root of 329 has a both a positive and a negative solution. Both 17 over two and the square root of 329 over two have the same denominator of two.

And we’ve just found that 𝑥 equals plus or minus the square root of 329 over two. And this is because the quadratic formula is an application of completing the squares. So both methods work. In a way, you could consider the quadratic formula a shortcut for completing the square. And both ways have the final answer of a solution set of 17.569 and negative 0.569.