Find the solution set of the equation five 𝑥 times 𝑥 minus six minus three times 𝑥 plus four plus three equals zero, giving values to one decimal place.
In this question, we’re given an equation which if we looked at it quickly, we might think it’s a linear equation, as it just has terms in 𝑥. However, if we look a little more closely, we notice that if we were to distribute this five 𝑥 across the parentheses, then we’d end up with five 𝑥 squared, meaning that this equation has a term in 𝑥 squared. And therefore, it’s a quadratic equation. So in order to find the solution set or solve to find the value of 𝑥, then it might be useful to put this quadratic equation into the general form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants.
We would start by distributing our values across the parentheses. So we’d end up with five 𝑥 squared minus 30𝑥 minus three 𝑥 minus 12 plus three is equal to zero. We can then collect the like terms, so we have five 𝑥 squared minus 33𝑥 minus nine is equal to zero. Now that we have our quadratic equation in this form, 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, we can solve it. There are a number of different ways we can solve the quadratic equation. For example, we could factor it. Notice however that this question asks us to give the values to one decimal place, and that’s usually a good clue that factoring won’t work. Therefore, another possible way in which we can find the solutions to this quadratic is by using the quadratic formula.
The quadratic formula tells us that if we have an equation in the general form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is not equal to zero, then 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. All we need to do is find the values of 𝑎, 𝑏, and 𝑐 in our quadratic equation and write those down ready to fill into the quadratic formula. 𝑎 is the coefficient of 𝑥 squared, which is five. 𝑏 is the coefficient of 𝑥, which is negative 33, and we need to make sure we’re including that negative sign. And 𝑐 is the constant at the end; it’s negative nine. So let’s write down the quadratic formula and fill in the values of 𝑎, 𝑏, and 𝑐.
When it comes to simplifying this equation, we can start with the fact that negative times negative 33 is simply 33. Then, within the square root, negative 33 squared is 1089. And then four times five is 20, times negative nine gives us negative 180. But we need to subtract this, and subtracting negative 180 is the same as adding 180. On the denominator, two times five gives us 10. We can then further simplify the value within the square root. 1089 plus 180 is 1269. Now we have that 𝑥 is equal to 33 plus or minus the square root of 1269 over 10. Remember that we use the plus or minus symbol in front of the square root to indicate that there are two solutions for 𝑥. 𝑥 is equal to 33 plus the square root of 1269 over 10, and 𝑥 is equal to 33 minus the square root of 1269 over 10.
In order to give the answer as a decimal, then we can pick up our calculators and input these calculations directly. Therefore, we have 𝑥 is equal to 6.862 and so on and 𝑥 is equal to negative 0.262 and so on. Remember that we need to give our values to one decimal place. When we check the second decimal digit of both of our values, we can see that both of these will be greater than five. So our answers are 𝑥 is equal to 6.9 and 𝑥 is equal to negative 0.3 to one decimal place. Usually, it’s fine to leave our answers in this form. But in this question, we were asked for the solution set. So we give our answer in set notation as the set containing 6.9 and negative 0.3.