Question

Find the zeros of the quadratic polynomial 5x^{2}-4-8x and verify the relationship between the zeros and the coefficient of the given polynomial.

Answer 1

5x^2-8x-4=0
5x^2-10x+2x-4=0
(x-2)(5x+2)=0
x=2 or x=-2/5
therefore zeroes are 2 and -2/5
now sum of zeroes = 2+(-2/5)=8/5
8/5 which is equal to -(coeff of x/coeff of x^2)
now
product of zeroes=2x(-2/5)=-4/5
-4/5 which is equal to (constant term/coeff of x^2)
Hence verified...

hope this helps:p

Answer 2

Let p(x)=5x^2-8x-4
=5x^2-10x+2x-4
=5x(x-2)+2(x-2)
=(x-2)(5x+2)
To find zeroes of we have to do p(x) =0
(x-2)(5x+2)=0
Therefore
x-2=0 or 5x+2=0
x=2 or x=-2/5
Let the zeroes are p and q
p=2 or q=-2/5
i) compare given p(x)=0 with ax^2+bx+c=0
a=5, b=-8, c=-4
i) sum of the zeroes =p+q=2-2/5=(10-2)/5=8/5

p+q= -b/a=-(-8)/5=8/5

ii) product of the zeroes= pq=2*(-2/5)=-4/5

pq=c/a=(-4)/5= -4/5