Question

Show that if k>3.2, the solutions of 5x²-8x+k=0 are not real numbers.

Answer 1

Answer:

3.2 is the highest value for k that will yield a non-negative discriminant

Step-by-step explanation:

For an equation to have real numbers as solutions, its discriminant (delta value) must be bigger than zero since there are no real square root solutions for negative numbers.

$x_{1} =\frac{-b +\sqrt{\Delta} }{2a}\\x_{2} =\frac{-b -\sqrt{\Delta} }{2a}$

In this case, the discriminant is given by:

$\Delta = (-8)^{2} - 4*5*k \\\Delta = 64 - 20*k$

For k =3.2

$\Delta = 64 - 64 = 0$

As shown above, 3.2 is the highest value for k that will yield a non-negative discriminant and, therefore, if k > 3.2 the solutions of the expression are not real numbers.