Question

The number of zeroes that polynomial f(x) = (x – 2)² + 4 can have is:​

Answer 1

Solution

_________________________

Given,

• (x-2)^2 +4

To find ,

• we have to find the number of zeros in the polynomial

Now,

• zero of the polynomial is the root of the polynomial

So,

• we have to first of the polynomial

$(x - 2) ^{2} + 4$

$= {x}^{2} + 4 - 4x + 4$

$= {x}^{2} - 4x + 8$

Now,

• solving the quadratic equation by discriminant method we get ;

we know that

$ax ^{2} + bx + c = 0$

so,

=> D = b^2 - 4ac

=> X = -b + √D / 2a

• compare the following equation with the general form we get ;

a=1 , b=-4, c= 8

Now,

$d \: = {b}^{2} - 4ac$

$= > d = ( - 4 )^{2} - 4(1)(8)$

$= > d = 16 - 32$

$= > d = - 16$

Answer

let the equation X = -b + √D / 2a

• solving this we get ;

=> X = -(-4)+√-16 / 2a

=> x = 4(+-)4a / 2

=> x = 2 (+-)2a

=> x = 2(1(+-) a)

Hence, zeros of the polynomial is 2(1+a) , 2(1-a)

_________________________

Answer 2

f(x) = (x – 2)² + 4

f(x) = x²+4-2x +4 = x²+8-2x

here no of degree is 2

hence it is quadrant polynomial

hence it have 2 zeroes