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

$f(x)\ =\ x^2\ –\ (\sqrt{3}\ +\ 1)x\ +\ \sqrt{3}$

Given:


$f(x) = x^2 – (\sqrt{3}+1)x +\sqrt{3}$

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,

$x^2 – (\sqrt{3}+1)x +\sqrt{3}= 0$

$x^2 – \sqrt{3}x - (1)x +\sqrt{3}= 0$

$x(x – \sqrt{3}) -1(x – \sqrt{3}) = 0$

$(x – \sqrt{3})(x -1) = 0$

$x-\sqrt{3}=0$ and $x-1=0$

$x = \sqrt{3}$ and $x = 1$

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

Verification:

We know that, 

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

                       $= –(\frac{-(\sqrt{3}+1)}{1})$

                       $=\sqrt{3}+1$

Sum of the zeros of $f(x)=\sqrt{3}+1$ 

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

                            $= \frac{\sqrt{3}}{1}$

                            $= \sqrt{3}$

Product of the roots of $f(x)=\sqrt{3}\times1 =\sqrt{3}$

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

Updated on: 10-Oct-2022

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