Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:

$f(x)\ =\ 6x^2\ –\ 3\ –\ 7x$

Given:


$f(x) = 6x^2 – 3-7x$

To find:

Here, we have to find the zeros of f(x). 

Solution:

To find the zeros of f(x), we have to put $f(x)=0$.

This implies,

$6x^2 – 3-7x = 0$

$6x^2 –7x-3 = 0$

$6x^2 – 9x +2x -3 = 0$

$3x(2x – 3) +1(2x– 3) = 0$

$(2x– 3)(3x+1) = 0$

$2x-3=0$ and $3x+1=0$

$2x= 3$ and $3x= -1$

$x=\frac{3}{2}$ and $x=\frac{-1}{3}$

Therefore, the zeros of the quadratic equation $f(x) = 6x^2 – 3 -7x$ are $\frac{3}{2}$ and $\frac{-1}{3}$.

Verification:

We know that, 

Sum of zeros $= -\frac{coefficient of x}{coefficient of x^2}$

                       $= –\frac{(-7)}{6}$

                       $=\frac{7}{6}$

Sum of the zeros of $f(x)=\frac{3}{2}+\frac{-1}{3}=\frac{3\times3-1\times2}{6}=\frac{9-2}{6}=\frac{7}{6}$

Product of roots $= \frac{constant}{coefficient of x^2}$

                            $= \frac{-3}{6}$

                           $=\frac{-1}{2}$

Product of the roots of $f(x)=\frac{3}{2}\times\frac{-1}{3} =\frac{-1}{2}$

Hence, the relationship between the zeros and their coefficients is verified.

Updated on: 10-Oct-2022

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