Question
Solve: $\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
✓
Answer
The given equation is $\text{x}^2-\big(\sqrt3+1\big)\text{x}+\sqrt3=0$
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=1,\ \text{b}=-\big(\sqrt3+1\big),\ \text{c}=\sqrt3$
$\therefore\text{D}=\text{b}^2-\text{4ac}$
$=\big[-\big(\sqrt3+1\big)\big]^2-4\times1\times\sqrt3$
$=\big(\sqrt3-1\big)^2>0$
So, the given equation has real roots, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]+\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1+\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]-\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1-\sqrt3-1}{2}$
$=\frac{2}{2}$
$=1$
Hence, $\sqrt3$ and $1$ are the roots of the given equation.
Comparing it with $ax^2+ bx + c = 0$, we get
$\text{a}=1,\ \text{b}=-\big(\sqrt3+1\big),\ \text{c}=\sqrt3$
$\therefore\text{D}=\text{b}^2-\text{4ac}$
$=\big[-\big(\sqrt3+1\big)\big]^2-4\times1\times\sqrt3$
$=\big(\sqrt3-1\big)^2>0$
So, the given equation has real roots, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]+\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1+\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]-\sqrt{\big(\sqrt3-1\big)^2}}{2\times1}$
$=\frac{\sqrt3+1-\sqrt3-1}{2}$
$=\frac{2}{2}$
$=1$
Hence, $\sqrt3$ and $1$ are the roots of the given equation.
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