### Video Transcript

Determine the type of the roots of the equation negative two π₯ minus six is equal to eight divided by π₯ plus seven.

In order to answer this question, we will firstly try and rewrite our equation so it is in the form ππ₯ squared plus ππ₯ plus π is equal to zero. We will begin by multiplying both sides of our equation by π₯ plus seven. On the left-hand side, we can then distribute the parentheses, otherwise known as expanding the brackets, using the FOIL method. Multiplying the first terms gives us negative two π₯ squared. The outer terms have a product of negative 14π₯. The inner terms multiply to give us negative six π₯, and the last terms multiply to give us negative 42. Our equation becomes negative two π₯ squared minus 14π₯ minus six π₯ minus 42 is equal to eight.

By subtracting eight from both sides of this equation and collecting like terms, we have negative two π₯ squared minus 20π₯ minus 50 is equal to zero. We can then divide through by negative two such that π₯ squared plus 10π₯ plus 25 equals zero. Our equation is now written in the form ππ₯ squared plus ππ₯ plus π equals zero, where π is equal to one, the coefficient of π₯ squared; π is equal to 10, the coefficient of π₯; and π, the constant, is equal to 25.

We can determine the type of roots in one of two ways. Firstly, we could solve the quadratic equation. One way of doing this would be by factoring, giving us π₯ plus five multiplied by π₯ plus five is equal to zero. As one of these parentheses must equal zero, we have two roots, π₯ equals negative five and π₯ equals negative five. This tells us that the two roots of the equation are real and equal.

An alternative method would be to find the discriminant. This is equal to π squared minus four ππ. If this value is greater than zero, we know that our roots are real and different. If the discriminant is equal to zero, our two roots are real and equal. Finally, if the value of the discriminant is less than zero, we know that we have complex roots. They are not real. Substituting in our values, we have 10 squared minus four multiplied by one multiplied by 25. This is equal to 100 minus 100. As this is equal to zero, we have confirmed that the roots are real and equal.

The equation negative two π₯ minus six is equal to eight divided by π₯ plus seven has two roots that are real and equal. They are equal to negative five.