Question

find the zeros of the polynomial and verify the relationship between the zeros and its coefficient,x^2+2√2x-8​

Answer 1

x²+2√2x-8=0

on comparing this eq with standard form of quadratic equation ax²+bx+c =0

we get ,

a= 1 , b= 2√2 and c= -8

D= b²-4ac = (2√2)² -4×-8 = 8+32 = 40

x= -b+-√D /2a = -2√2 +- √40 /2 = -2.82 +-6.32 /2

x = -3.5 /2 and x= -9.14/2

Answer 2

Step-by-step explanation:

ANSWER

f(x)=x

2

−2x−8

⇒f(x)=x

2

−4x+2x−8

⇒f(x)=x(x−4)+2(x−4)]

⇒f(x)=(x−4)(x+2)

Zeros of f(x) are given by f(x) = 0

⇒x

2

−2x−8=0

⇒(x−4)(x+2)=0

⇒x=4 or x=−2

So, α=4 and β=−2

∴ sum of zeros =α+β=4−2=2

Also, sum of zeros =

Coefficient of x

2

Coefficient of x

=

1

−(−2)

=2

So, sum of zeros =α+β=−

Coefficient ofx

2

Coefficient of x

Now, product of zeros =αβ=(4)(−2)=−8

Also, product of zeros =

Coefficient ofx

2

Constant term

=

1

−8

=−8

∴ Product of zeros =

Coefficient of x

2

Constant term

=αβ