<s>[INST] Cosine rule problem Question:
In triangle ABC, BC = a, AC= b, AB = c, and BD is perpendicular to AC.
If $\angle ABC = 60^{\circ}$, prove that $c = \dfrac{1}{2}a \pm \sqrt{b^2-\dfrac{3}{4}a^2}$
My approach: $b^2 = a^2 + c^2 - 2ac\cos B$
$b^2 = a^2 + c^2 - 2ac \cos{60}^\circ$
$b^2 = a^2 + c^2 - ac$
$b^2 = (a-c)^2+ac$
After this I don't know how to isolate the $c$ term.
 [/INST] Your work is correct. Picking up your work from $b^2=a^2+c^2-ac$, subtract $b^2$ from both sides and rewrite to get
$$c^2-ac+a^2-b^2=0$$
which is quadratic in $c$. Notice this equation fits the standard quadratic form with $A=1$, $B=-a$ and $C=a^2-b^2$.
Using the quadratic formula, we get
$$c=\frac{a\pm\sqrt{a^2-4(a^2-b^2)}}{2}$$
$$c=\frac{a}{2}\pm\frac{\sqrt{4b^2-3a^2}}{2}$$
$$c=\frac{a}{2}\pm\sqrt{b^2-\frac{3}{4}a^2}\quad.$$
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