Find the solution set of the equation 2𝑥 − 5 = 6/𝑥, giving values to three decimal places.

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### Video Transcript

Find the solution set of the equation two 𝑥 minus five equals six over 𝑥, giving values to three decimal places.

Now, at first glance, we may not know how to solve this equation as it involves a reciprocal term, six over 𝑥. There’s a trick, though, that we need to spot, which is that if we multiply the entire equation by 𝑥, this will eliminate the fraction. On the left-hand side, we’ll have the 𝑥 multiplied by two 𝑥 minus five. And on the right-hand side, 𝑥 multiplied by six over 𝑥. We can distribute the parentheses or expand the brackets on the left-hand side to give two 𝑥 squared minus five 𝑥. And on the right-hand side, 𝑥 multiplied by six over 𝑥 simplifies to six. And we’ve eliminated the fraction. We now see that what we have is a quadratic equation. And so, our next step is to collect all the terms on the same side.

We can do this by subtracting six from each side of the equation, giving two 𝑥 squared minus five 𝑥 minus six is equal to zero, a quadratic equation in its most easily recognizable form. Now, we could perform a quick check to see whether this quadratic equation can be solved by factoring. But as we’re told in the question to give the solution set to three decimal places, that’s a huge clue that it can’t. So, we’re going to need to apply the quadratic formula instead. This tells us that, for the general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. Then the solution set or the roots of this quadratic equation are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

Now, it’s important to notice that the quadratic formula uses a quadratic equation in the form where all the terms are on one side, and we have zero on the other. So, when we’re identifying the values of 𝑎, 𝑏, and 𝑐 to substitute into the formula, we must use our quadratic equation in its rearranged form, not in its earlier form of two 𝑥 squared minus five 𝑥 equals six. If we were to use the earlier form, then we would have the sign for the value of 𝑐 incorrect, which would lead to an incorrect solution. Let’s determine then the values of 𝑎, 𝑏, and 𝑐 for our quadratic equation. 𝑎 is the coefficient of 𝑥 squared. That’s equal to two. 𝑏 is the coefficient of 𝑥, which is equal to negative five. And finally, 𝑐 is the constant term, which is equal to negative six. Remember, we must include the sign with each of these values.

Next, we need to substitute the values of 𝑎, 𝑏, and 𝑐 carefully into the quadratic formula. We have 𝑥 equals negative 𝑏 — that’s negative negative five — plus or minus the square root of 𝑏 squared, negative five squared, minus four 𝑎𝑐 — that’s minus four times two times negative six — all over two 𝑎, which is two times two. Now, we simplify, being very careful with the negative signs. Negative negative five is simply five. And in the denominator, two times two is four. Within the square root, negative five squared is 25.

Be very careful here, particularly if you’re using a calculator. A common mistake is to write that as negative 25. We’re then subtracting four times two times negative six. So, we’re subtracting negative 48. Within the square root, 25 minus negative 48 is the same as 25 plus 48, which is 73. So, we have that 𝑥 is equal to five plus or minus the square root of 73 over four. And that’s our exact answer in terms of surds.

The question though asked us to give our values to three decimal places. So, we now need to evaluate these on a calculator. Our first value is 𝑥 equals five plus the square root of 73 all over four, which is 3.38600. Our second value is five minus the square root of 73 all over four, which is negative 0.88600. Rounding these values to three decimal places then gives the solution set of 3.386, negative 0.886.