Question

If one root of the equation x² $-$ 10x + 2k = 0 is 5 $-$ $\sqrt{3}$, find k.

Answer 1

If one root of the quadratic equation is $5-\sqrt{3}$, then the other root of the quadratic equation is its conjugate, that is, $5+\sqrt{3}$
Thus,

$$\begin{gathered} \mathrm{\alpha }=5-\sqrt{3}\mathrm{and}\mathrm{\beta }=5+\sqrt{3} \\ \mathrm{Now}, \\ \mathrm{\alpha }+\mathrm{\beta }=5-\sqrt{3}+5+\sqrt{3} \\ =10 \\ \mathrm{\alpha \beta }=(5-\sqrt{3})(5+\sqrt{3}) \\ ={\left(5\right)}^2-{\left(\sqrt{3}\right)}^2 \\ =25-3 \\ =22 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{\alpha }+\mathrm{\beta }=\frac{-b}{a}\mathrm{and}\mathrm{\alpha \beta }=\frac{c}{a}. \end{gathered}$$

Now, on comparing the equation with ax² + bx + c = 0 in variable x , we get:
a = 1, b = –10 and c = 2k
From $\mathrm{\alpha \beta }=\frac{c}{a}$ , we have:
$22=\frac{2k}{1}=>k=\frac{22}{2}=11$