MCQ
If two zeros $x^3+x^2-5 x-5$ are $\sqrt{5}$ and $-\sqrt{5},$ then its third zero is :
- A$1$
- ✓$-1$
- C$2$
- D$-2$
✓
Answer
Correct option: B.
$-1$
Let $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ be the given zeros and $\gamma$ be the third zero of $x^3+x^2-5 x-5=0$
By using $\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}$
$\alpha+\beta+\gamma=-\frac{+(+1)}{1}$
$\alpha+\beta+\gamma=-1$
By substituting $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ in $\alpha+\beta+\gamma=-1$
$\sqrt{5}-\sqrt{5}+\gamma=-1$
$\gamma=-1$
Hence, the correct choice is $(b)$
By using $\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}$
$\alpha+\beta+\gamma=-\frac{+(+1)}{1}$
$\alpha+\beta+\gamma=-1$
By substituting $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ in $\alpha+\beta+\gamma=-1$
$\sqrt{5}-\sqrt{5}+\gamma=-1$
$\gamma=-1$
Hence, the correct choice is $(b)$
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