Question

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and their coefficients.

$48y^2-13y-1$

• A. $\frac{1}{7},\frac{-1}{16}$
• B. $\frac{1}{2},\frac{-1}{14}$
• C. $\frac{1}{2},\frac{-1}{11}$
• D. $\frac{1}{3},\frac{-1}{16}$

Answer 1

The correct option is D $\frac{1}{3},\frac{-1}{16}$

$48y^2-13y-1$

$=48y^2-16y+3y-1$

$=16y(3y-1)+1(3y-1)$

$=(3y-1)(16y+1)$

$x=\frac{1}{3}$ , $x=-\frac{1}{16}$

The zeros of the polynomials are $\frac{1}{3}$ , $-\frac{1}{16}$

Relationship between the zeros and the coefficients of the polynomials-

Sum of the zeros=-$\frac{coefficientofy}{coefficientof{y}^2}=-\left(\frac{-13}{48}\right)=\frac{13}{48}$

Also sum of zeros=$\frac{1}{3}-\frac{1}{16}$ =$\frac{16-3}{48}$ =$\frac{13}{48}$

Product of the zeros =$\frac{constantterm}{coefficientof{y}^2}=\frac{-1}{48}$

Also the product of the zeros= $\left(\frac{1}{3}\right)\times (-\frac{1}{16})=\frac{-1}{48}$

Hence verified.