Find the solution set of 𝑥 plus 17 over 𝑥 is equal to six in the set of real values, giving values to one decimal place.
We recall that we can solve any quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is not equal to zero, using the quadratic formula. This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎.
Our first step in this question is to rearrange our equation 𝑥 plus 17 over 𝑥 is equal to six so it is in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. We begin by multiplying each of our terms by 𝑥. 𝑥 multiplied by 𝑥 is equal to 𝑥 squared. 17 over 𝑥 multiplied by 𝑥 is equal to 17. And on the right-hand side, six multiplied by 𝑥 is six 𝑥.
Next, we can subtract six 𝑥 from both sides of the equation. This leaves us with the quadratic equation 𝑥 squared minus six 𝑥 plus 17 is equal to zero. The constants 𝑎, 𝑏, and 𝑐 are equal to one, negative six, and 17, respectively. Substituting these values into the quadratic formula gives us 𝑥 is equal to negative negative six plus or minus the square root of negative six squared minus four multiplied by one multiplied by 17 all divided by two multiplied by one. Squaring negative six gives us 36. Multiplying four, one, and 17 gives us 68. Two multiplied by one is equal to two. 𝑥 is therefore equal to six plus or minus the square root of 36 minus 68 all divided by two. Underneath the square root, 36 minus 68 is equal to negative 32.
We were asked to give solutions in the set of real numbers. And we know that the square root of any negative number gives no real solutions. This means that there are no real solutions to the square root of negative 32. This means that the equation 𝑥 plus 17 over 𝑥 equals six has no solutions in the set of real numbers. The correct answer is therefore the empty set as shown.