Question

find a quadratic equation whose zeroes are 5 and -4 respectively

Answer 1

Let the zeroes be x and y

Now According To Question

x+y=-5

xy=-4

Now Polynomial is a^2-(x+y)a+xy

Polynomial Is :a^2-(-5)a+(-4)

a^2+5a-4

Solving The Equation :-

a^2+1a-4a-4=0

a(a+1)-4(a+1)=0

(a+1)(a-4)=0

a = -1 or 4

Answer 2

Answer:

1x2 - 1x - 20

Step-by-step explanation:

sum of zeroes = -b/a = $\alpha +\beta$=5-4=1

product of zeroes = c/a=$\alpha \beta$= 5*(-4)=-20

therefore,

a=1

b=-1

c=-20

thus, the quadratic equation is

1x2 - 1x- 20