Question

If $\alpha ,\beta$ and $\gamma$ are real numbers such that ${\alpha }^2+{\beta }^2+{\gamma }^2=1$ and $\alpha +\beta +\gamma =\sqrt{3},$ then $\beta =$
(correct answer + 1, wrong answer - 0.25)

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{4}{\sqrt{3}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\pm \frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

${\alpha }^2+{\beta }^2+{\gamma }^2=1\cdots \left(i\right)$
$\alpha +\beta +\gamma =\sqrt{3}\cdots \left(ii\right)$
From $\left(ii\right)$, we have $\gamma =\sqrt{3}-\alpha -\beta \cdots \left(iii\right)$.
Substituting $\left(iii\right)$ into $\left(i\right)$:
${\alpha }^2+{\beta }^2+(\sqrt{3}-\alpha -\beta {)}^2=1$
$\Rightarrow {\alpha }^2+{\beta }^2+3+{\alpha }^2+{\beta }^2-2\sqrt{3}\alpha -2\sqrt{3}\beta +2\alpha \beta =1$
$\Rightarrow {\alpha }^2+(\beta -\sqrt{3})\alpha +({\beta }^2-\sqrt{3}\beta +1)=0$

Since, $\alpha$ is real, $D\ge 0$
$\Rightarrow (\beta -\sqrt{3}{)}^2-4({\beta }^2-\sqrt{3}\beta +1)\ge 0$
$\Rightarrow (\sqrt{3}\beta -1{)}^2\le 0$
$\Rightarrow \beta =\frac{1}{\sqrt{3}}$

Alternate Solution:
${\alpha }^2+{\beta }^2+{\gamma }^2=1$ and $\alpha +\beta +\gamma =\sqrt{3}$
Let $\alpha =\beta =\gamma$
$\Rightarrow 3{\alpha }^2=1$
$\Rightarrow \alpha =\pm \frac{1}{\sqrt{3}}...\left(1\right)$
And $3\alpha =\sqrt{3}$
$\Rightarrow \alpha =\frac{1}{\sqrt{3}}...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$\alpha =\beta =\gamma =\frac{1}{\sqrt{3}}$