# Question Video: Solving Quadratic Equations Using the Quadratic Formula Mathematics

Find the solution set of the equation (3/(𝑥 + 3)) − (4/(𝑥 − 3)) = 3 in ℝ, giving values to one decimal place.

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### Video Transcript

Find the solution set of the equation three over 𝑥 plus three minus four over 𝑥 minus three is equal to three in the set of real values, giving values to one decimal place.

It might not be immediately obvious how we can solve the equation in this question. One strategy for an equation of this type is to try and eliminate the denominators by firstly finding the lowest common denominator. When adding one-third and one-quarter, the easiest way to find a common denominator is to multiply the two denominators together. Three multiplied by four is equal to 12. Therefore, our common denominator is equal to 12. Whilst our equation is more complicated, we can use the same method. We have denominators of 𝑥 plus three and 𝑥 minus three. Multiplying these together, we have 𝑥 plus three multiplied by 𝑥 minus three. By multiplying each of the three terms in our equation by this, we will be able to eliminate the denominators.

Multiplying the first term by 𝑥 plus three and 𝑥 minus three gives us three 𝑥 plus three 𝑥 minus three divided by 𝑥 plus three. This simplifies to three multiplied by 𝑥 minus three. The second term becomes negative four multiplied by 𝑥 plus three multiplied by 𝑥 minus three divided by 𝑥 minus three. This time we can cancel a factor of 𝑥 minus three from the numerator and denominator, leaving us with negative four multiplied by 𝑥 plus three. On the right-hand side, we have three multiplied by 𝑥 plus three multiplied by 𝑥 minus three.

At this stage, we might notice that 𝑥 plus three multiplied by 𝑥 minus three is the factorization of the difference of two squares. We know that 𝑥 squared minus 𝑎 squared is equal to 𝑥 plus 𝑎 multiplied by 𝑥 minus 𝑎. This means that 𝑥 plus three multiplied by 𝑥 minus three is equal to 𝑥 squared minus nine. The right-hand side of our equation becomes three multiplied by 𝑥 squared minus nine.

Our next step is to distribute the parentheses or expand the brackets. This gives us three 𝑥 minus nine. On the left-hand side, we also have negative four 𝑥 minus 12. And on the right-hand side, we have three 𝑥 squared minus 27. The left-hand side simplifies to negative 𝑥 minus 21. We can then add 𝑥 and 21 to both sides of the equation such that three 𝑥 squared plus 𝑥 minus six equal zero. This is a quadratic equation written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero.

We know that one way to solve an equation of this type, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, is 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. If the value of 𝑏 squared minus four 𝑎𝑐, known as the discriminant, is greater than or equal to naught, the equation will have real solutions. In our equation, the values of 𝑎, 𝑏, and 𝑐 are three, one, and negative six, respectively. This means that 𝑥 is equal to negative one plus or minus the square root of one squared minus four multiplied by three multiplied by negative six all divided by two multiplied by three. This simplifies to negative one plus or minus the square root of 73 all divided by six.

We have two possible solutions either 𝑥 is equal to negative one plus the square root of 73 divided by six or 𝑥 is equal to negative one minus the square root of 73 divided by six. Typing these into the calculator, we have 𝑥 is equal to 1.257 and so on or 𝑥 is equal to negative 1.590 and so on. We have been asked to give our solutions to one decimal place. The solution set of the equation three over 𝑥 plus three minus four over 𝑥 minus three equals three contains the values 1.3 and negative 1.6 correct to one decimal place.

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