Question

Please solve the problem

Answer 1

Answer:

u just check the question of 5.

Step-by-step explanation:

1. p(x)=9x^2 - 17x

let

9x^2-17x=0

x(9x-17)=0

9x-17=0

9x=17

x=17/9

so 17/9 is zero of polynomial p(x)

  2. p(x)=5x^2+7x

let,

5x^2+7x=0

x(5x+7)=0

5x=-7

x=-7/5

so -7/5 is the zero of the polynomial p(x)

3. p(x)=36x^2-25

let,

36x^2-25=0

(6x)^2-(5)^2=0

(6x-5)(6x+5)=0

x=5/6 , -5/6  

so 5/6 and -5/6 are the zeroes of the polynomial p(x)

4. p(x)=2x^2-9

let,

2x^2-9=0

(root2x)^2-3^2=0

(root2x-3)(root2x+3)=0

x=3/root2 or -3/root2

so 3/root2 or -3/root2 are zeroes of the polynomial p(x)

5. p(x)=4x^2+9

let,

4x^2+9=0

4x^2=-9

x^2=-9/4

x=root of -9/4 which is 3/2 * iota (its an imaginary no. whose value is root -1.)

                                  ........end.....