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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$q(x)\ =\ \sqrt{3}x^2\ +\ 10x\ +\ 7\sqrt{3}$
Given:
$f(x) = \sqrt{3}x^2 + 10x + 7\sqrt{3}$
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,
$\sqrt{3}x^2 + 10x + 7\sqrt{3}= 0$
$\sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0$
$\sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0$
$(x + \sqrt{3})(\sqrt{3}x + 7) = 0$
$x+\sqrt{3}=0$ and $\sqrt{3}x+7=0$
$x = -\sqrt{3}$ and $\sqrt{3}x = -7$
$x = -\sqrt{3}$ and $x=\frac{-7}{\sqrt{3}}$
Therefore, the zeros of the quadratic equation $f(x) = \sqrt{3}x^2 + 10x + 7\sqrt{3}$ are $-\sqrt{3}$ and $\frac{-7}{\sqrt{3}}$.
Verification:
We know that,
Sum of zeros $= -\frac{coefficient of x}{coefficient of x^2}$
$= –\frac{10}{\sqrt{3}}$
Sum of the zeros of $f(x)=-\sqrt{3}+(\frac{-7}{\sqrt{3}})=\frac{-\sqrt{3}\times\sqrt{3}+(-7)}{\sqrt{3}}=\frac{-3-7}{\sqrt{3}}=\frac{-10}{\sqrt{3}}$
Product of roots $= \frac{constant}{coefficient of x^2}$
$= \frac{7\sqrt{3}}{\sqrt{3}}$
$= 7$
Product of the roots of $f(x)=-\sqrt{3}\times\frac{-7}{\sqrt{3}}=7$
Hence, the relationship between the zeros and their coefficients is verified.
Updated on: 10-Oct-2022
11K+ Views
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