For what value of k are there two distinct real solutions to the original quadratic equation (k+1)x²+4kx+2=0.

k ∈ (-∞,)∪(1,∞)

Step-by-step explanation:

For quadratic equations you can find the solutions with the Bhaskara's Formula:

A quadratic equation usually has two solutions.

If you only want real solutions the condition is that the discriminant () has to be greater than zero, this means:

Then we have the expression:

Now to find two distinct real solutions to the original quadratic equation we have to calculate the discriminant:

We got another quadratic function.

we can simplify the expression dividing both sides in 8.

We can apply Bhaskara's Formula except that the condition in this case is that the solutions have to be greater than zero.

Then,

The answer is:

For all the real values of k who belongs to the interval:

(-∞,)∪(1,∞)

there are two distinct real solutions to the original quadratic equation

tangent

step-by-step explanation: