Homework Question

Question

what number could you put in the blank to create an equation that has two imaginary solutions?

x^{2}-4x-5=______

Answer 1

Based on the left-hand side...

x² - 4x + 5 --> (x-1)*(x+5)

So we see that there are two real solutions currently...

However, we can let our answer to find be represented as C and use the quadratic formula to find out bounds for U...

x² - 4x + (5 - U) = 0

Define A = coefficient on x² --> 1

Define B = coefficient on x --> -4

Define C = constant --> 5 - U

x = [ -B +/- sqrt(B² - 4AC) ] / (2A)

We need B² - 4AC < 0 to make an imaginary number...

Solving for U in this inequality...

(-4)² - 4*1*(5-U) < 0

16 - 20 + 4U < 0

-4 + 4U < 0

4*( -1 + U) < 0

-1 + U < 0

-1 < -U

U > 1

U > 1 --> so we have a range of values for U where this will cause the discriminant (B² - 4AC) to be negative before the sqrt root is taken causing complex imaginary solutions.