Question Video: Solving Quadratic Equations by Expanding and Using the Quadratic Formula Mathematics

Find the solution set of the equation 3𝑥² + 4(𝑥 + 1) = 0, giving values in ℝ to one decimal place.

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Video Transcript

Find the solution set of the equation three 𝑥 squared plus four multiplied by 𝑥 plus one equals zero, giving values in the set of real numbers to one decimal place.

In order to answer this question, we firstly need to rearrange our equation so it is in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. This will enable us to use the quadratic formula to solve it. Distributing the parentheses by multiplying four by 𝑥 and four by one gives us three 𝑥 squared plus four 𝑥 plus four equals zero. Our values of 𝑎, 𝑏, and 𝑐 are three, four, and four, respectively. 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.

Substituting in our values, we have 𝑥 is equal to negative four plus or minus the square root of four squared minus four multiplied by three multiplied by four all divided by two multiplied by three. Four squared is 16, four multiplied by three multiplied by four is 48, and two multiplied by three is six. As 16 minus 48 is negative 32, we are left with 𝑥 is equal to negative four plus or minus the square root of negative 32 all divided by six.

At this stage, as we want solutions to one decimal place, we would usually input our calculation into our calculator. However, this calculation involves taking the square root of a negative number, negative 32. And we know that square rooting a negative number has no real solutions. And when we type it into the calculator, we get a mathematical error. This means that there are no real solutions to our equation. And the solution set of the equation is the empty set. This leads us to a key fact about the quadratic formula. If the expression underneath the square root 𝑏 squared minus four 𝑎𝑐, known as the discriminant, is less than zero, then there are no real solutions to our quadratic equation.

It is also worth considering what this would look like graphically. The quadratic equation 𝑦 is equal to three 𝑥 squared plus four multiplied by 𝑥 plus one is shown in the figure. We notice that the graph does not intersect the 𝑥-axis. This confirms that there are no real solutions to the equation. Any quadratic equation, where the discriminant 𝑏 squared minus four 𝑎𝑐 is less than zero, will not intersect the 𝑥-axis.

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