Find values of b so that t^(2)+bt+81=0 has one real solution

Find the values of b so that t^(2)+bt+81=0 has one real solution.

I have been having trouble doing this so it would be most appreciated if I received help sooner then later.

HelpMePlease123 said:

Find the values of b so that t^(2)+bt+81=0 has one real solution.

I have been having trouble doing this so it would be most appreciated if I received help sooner then later.

Click to expand...

This is a quadratic equation!

How many real solutions can it have?

What will be the condition under which it will have ONE real solution?

HelpMePlease123 said:

Find the values of b so that t^(2)+bt+81=0 has one real solution.

I have been having trouble doing this so it would be most appreciated if I received help sooner then later.

Click to expand...

Depending on how much you have learned about quadratic equations, there are several ways you might approach this.

One is to use the discriminant, or equivalently the quadratic formula. What can that tell you about the number of real solutions?

If you don't know about that, you could think about what it would look like in factored form if it has one solution, and then find what the factors would be. This would determine b.

What have you learned, and what ideas do you have about it?

HelpMePlease123 said:

Find the values of b so that t^(2)+bt+81=0 has one real solution.

I have been having trouble doing this so it would be most appreciated if I received help sooner then later.

Click to expand...

The Quadratic Formula can give two solutions, because of the "plus/minus" on the square root. What would need to be true of the contents of that square root, in order for you to get only and exactly one real-number solution to the equation?