Question

Jester 2

Answer 1

Answer:

$c < \frac{9}{8}$

For all c less than $\frac{9}{8}$ ,the given quadratic equation has no real number solutions.

Step-by-step explanation:

Given

Quadratic Equation

$-2x^{2} + 3x + c = 0$

On comparing with general form of Quadratic Equation

$Ax^{2}+ Bx + C = 0$

we have,

$A=-2\\B=3\\C=c$

For no real number solution we must have

$Discriminant, B^{2}- 4AC < 0\\(3)^{2} - 4\times -2\times c < 0\\9-8c < 0\\-8c < -9\\c < \frac{9}{8}$

So, For all c less than $\frac{9}{8}$ ,the given quadratic equation has no real number solutions.