Question

7. Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j2 + 3j – 28 = 0.

Answer 1

`\bf \begin{array}{llccll} y=&{{ -4}}j^2&{{ +3}}j&{{ -28}}\\ &~\uparrow &\uparrow &\uparrow \\ &~a&b&c \end{array} \\\\\\ discriminant\implies b^2-4ac= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases}`

so.. check what value the discriminant is then.

Answer 2

Answer with Step-by-step explanation:

We know that the discriminant(d) of a quadratic equation ax²+bx+c=0 is given by:

d=b²-4ac

Also, if d=0 then we have two equal real roots

if d is positive i.e. d>0 then, we have two unequal real roots

and when d is negative i.e. d<0 then, we have no real root

Here, We are given equation -4j²+3j-28

a= -4

b=3

and c= -28

So, d=(3)²- 4×(-4)×(-28)

d= 9-448

d= -439

d<0

Hence, equation -4j²+3j-28 has no real roots