Question Video: Solving Quadratic Equations Using the Quadratic Formula | Nagwa Question Video: Solving Quadratic Equations Using the Quadratic Formula | Nagwa

Question Video: Solving Quadratic Equations Using the Quadratic Formula Mathematics

Find the solution set of the equation (18/𝑥²) + (5/𝑥) = 1, giving values to three decimal places.

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Video Transcript

Find the solution set of the equation 18 over 𝑥 squared plus five over 𝑥 equals one, giving values to three decimal places.

It may not immediately be obvious how we can solve an equation of this type. However, a useful strategy to simplify the equation will be to try and eliminate the denominators. One way of doing this is to multiply through by the lowest common denominator. In this question, we have denominators of 𝑥 squared and 𝑥. The lowest common denominator here is 𝑥 squared, which means we can multiply both sides of our equation by 𝑥 squared.

Multiplying 𝑥 squared by 18 over 𝑥 squared gives us 18𝑥 squared over 𝑥 squared. This simplifies to 18. In a similar way, when we multiply 𝑥 squared by five over 𝑥, we get five 𝑥 squared over 𝑥. Canceling an 𝑥 here, this simplifies to five 𝑥. Multiplying the left-hand side of our equation by 𝑥 squared gives us 18 plus five 𝑥. And this is equal to 𝑥 squared.

We now have a quadratic equation that we can simplify by rewriting all of the terms on one side of the equals sign. We could subtract 𝑥 squared from both sides or alternatively subtract 18 and five 𝑥, giving us 𝑥 squared minus five 𝑥 minus 18 equals zero. This is a quadratic equation written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero.

We know that one way of solving an equation of this type is to use the quadratic formula. When 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, the quadratic formula 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. In our equation, 𝑎 is equal to one, 𝑏 is negative five, and 𝑐 is negative 18. Substituting in these values, we have 𝑥 is equal to negative negative five plus or minus the square root of negative five squared minus four multiplied by one multiplied by negative 18 all divided by two multiplied by one.

At this stage, it is worth recalling that the value of 𝑏 squared minus four 𝑎𝑐, known as the discriminant, must be greater than or equal to naught for us to have any real solutions. It is also important to note that when we square any negative number on the calculator, we must put the negative number in parentheses or brackets. Simply typing negative five squared into the calculator gives us negative 25. However, we know that squaring a negative number, in this case, negative five, gives a positive answer.

Simplifying the right-hand side of our equation gives us five plus or minus the square root of 97 all divided by two. This means that either 𝑥 is equal to five plus the square root of 97 all divided by two or 𝑥 is equal to five minus the square root of 97 all divided by two. Typing these into our calculator gives us 𝑥 is equal to 7.4244 and so on and 𝑥 is equal to negative 2.4244 and so on.

We are asked to round our solutions to three decimal places. The solution set of the equation 18 over 𝑥 squared plus five over 𝑥 equals one contains the values 7.424 and negative 2.424, correct to three decimal places.

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