Question

zeroes of a quadratic polynomial p(x) are reciprocal of zeroes of q(x)=6x²-5x+1​

Answer 1

Given: factor of p(x) ?are reciprocal of q(x)

factors of q(x):6x^2-5x+1

6x^2-6x+x+1

6x(x-1)-1(x-1)

(6x-1)(x-1)=0

x=1/6;x=1

Now reciprocal of factors :

factors of p(x)=> x=6 /1 ; x= 1/1

let a= 6 and b=1

therefore,

using the formula: x^2-(a+b)x+(ab)

=> x^2- (6+1)x+(6×1)

=>x^2-7x+6(ans)

now it is correct

Answer 2

Answer:

Step-by-step explanation:

q(x) =6x^2 - 5x-1

=6x^2 - 6x+x-1

=6x(x-1)+1(x-1)

=(6x+1)(x-1)

So - 1/6&1are the zeroes of q(x).

Therefore.,ATQ

Zeroes of p(x) =(-6) & 1

Let alpha=(-6) & beta=1

So p(x)

=x^2 - (Alpha +beta)x +(alpha ×beta)

=x^2 - (-6+1)x +(-6×1)

=x^2 +5x-6(ans).