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

$f (v) = v^{2} + 4 \sqrt{3} v - 15$

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\sqrt{2}

and

5 \sqrt{3}

- \sqrt{3}

and

- 4 \sqrt{2}

5 \sqrt{3}

and

5 \sqrt{2}

\sqrt{3}

and

- 5 \sqrt{3}

answer is D.

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Detailed Solution

Given, quadratic function,

f (v) = v^{2} + 4 \sqrt{3} v - 15

We put the value

f (v) = 0

\Rightarrow v^{2} - \sqrt{3} v + 5 \sqrt{3} v - 15 = 0

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

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

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

v = \sqrt{3}, - 5 \sqrt{3}

Thus, zeros of the polynomial

v = \sqrt{3}, - 5 \sqrt{3}

For the verification of the relation between 0 and their coefficients
Using this formula,

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

\sqrt{3} + (- 5 \sqrt{3}) = - \frac{4 \sqrt{3}}{1}

- 4 \sqrt{3} = - 4 \sqrt{3}

We calculate

Product of roots

= \frac{constant}{coefficient of v^{2}}

\sqrt{3} \times (- 5 \sqrt{3}) = \frac{- 15}{1}

- 5 \times 3 = - 15

- 15 = - 15

Verified the relation between zeros and their coefficients.
Hence, the zeros of each of

f (v) = v^{2} + 4 \sqrt{3} v - 15

v = \sqrt{3}, - 5 \sqrt{3}

Option 4 is correct.

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