For the equation 5×2+px+5=0,p>0 if one root of the equation is square of the other, then p=

# For the equation $5{x}^{2}+px+5=0,p>0$ if one root of the equation is square of the other, then $p=$

1. A

$-10$

2. B

$5$

3. C

$-5$

4. D

Both $a$ and  $b$

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### Solution:

$\alpha ,{\alpha }^{2}$ are the roots of  $5{x}^{2}+px+5=0$

$\alpha +{\alpha }^{2}=\frac{-p}{5}$ and $\alpha .{\alpha }^{2}=\frac{5}{5}=1$

$⇒{\alpha }^{3}=1$

$⇒\alpha =1.\omega ,{\omega }^{2}$

$1+1=\frac{-p}{5}$

$⇒p=-10$

$\alpha =\omega ,\omega +{\omega }^{2}=\frac{-p}{5}⇒-1=\frac{-p}{5}⇒p=5$

$\alpha ={\omega }^{2},{\omega }^{2}+{\omega }^{4}=\frac{-p}{5}⇒{\omega }^{2}+\omega =\frac{-p}{5}⇒p=5$

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