Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$f(v)\ =\ v^2\ +\ 4\sqrt{3}v\ –\ 15$

Given:


$f(v) = v^2+4\sqrt{3}v – 15$

To find:

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

Solution:

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

This implies,

$v^2+4\sqrt{3}v – 15= 0$

$v^2+5\sqrt{3}v -\sqrt{3}v– 15= 0$

$v(v+5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})= 0$

$(v+5\sqrt{3})(v-\sqrt{3}) = 0$

$v+5\sqrt{3}=0$ and $v-\sqrt{3}=0$

$v = -5\sqrt{3}$ and $v = \sqrt{3}$

Therefore, the zeros of the quadratic equation $f(v) = v^2+4\sqrt{3}v – 15$ are $-5\sqrt{3}$ and $\sqrt{3}$.

Verification:

We know that, 
Sum of zeros $= -\frac{coefficient of v}{coefficient of v^2}$

                       $= –\frac{4\sqrt{3}}{1}$

                       $=-4\sqrt{3}$

Sum of the zeros of $f(v)=-5\sqrt{3}+\sqrt{3}=-4\sqrt{3}$.

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

                            $=\frac{-15}{1}$

                            $= -15$

Product of the roots of $f(v)=-5\sqrt{3}\times \sqrt{3}=-15$.

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

Updated on: 10-Oct-2022

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