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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$f(x)\ =\ x^2\ β\ (\sqrt{3}\ +\ 1)x\ +\ \sqrt{3}$
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.
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