# Solving triangles and quadratic equations

When calculating the pieces in a triangle with only two sides and an intermediate angle is known, one must solve a quadratic equation. By solving the equation are 2, 1 or 0 solutions, as described below (A-D). These four options is illustrated in Figure 1-4 below. You must write at the bottom of the options (A-D) belongs to the illustrations (1-4)

A: there should be 2 solutions (both positive)

B: there must be 2 solutions (one positive and one negative)

C: there must be a solution

D: must be 0 solutions

Problem is that I have no idea how to start or how to find solutions. I hope anyone who wants to 'make one so I can follow them and do the rest.

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You can read the answers for these particular choices of numbers off the pictures. Are you expected to set up suitable quadratic equations and solve them? –  André Nicolas Feb 5 '12 at 19:45
The first one got 0 solution ? what about 2, 3 and 4 ? can you also read them without quadratic equations ? that some power you have –  user1022734 Feb 5 '12 at 21:54
Yes, the first one has no solution. –  André Nicolas Feb 5 '12 at 21:57
What about two, three and four? two got 2 solution , one positive one negative ? right –  user1022734 Feb 5 '12 at 21:59
For 2) probably two positive solutions, it really depends on how you are setting up your equation, and we are not told that in the post. For 3), it will probably turn out to be one positive and one negative, the negative to be discarded. For 4) it will definitely be one solution. –  André Nicolas Feb 5 '12 at 22:08

If you can trust the pictures, then no calculation is necessary. The first picture, for example, depicts a "side-side-angle" setup in which no choice of the angle $C$ will yield a valid triangle. If you want to verify this numerically, then probably you should check that one of the relations that holds in triangles fails.
Here law of cosines might help: if there were such a triangle and hence a valid choice for $c$, the length of the side opposite $C$, then we'd have \begin{align*} 5^2 &= 13^2 + c^2 - 2 \cdot 13c\,\cos29^\circ \\ 0 &= c^2 - (26\,\cos29^\circ)c + 144 \end{align*} Look at the discriminant: $(26\,\cos29^\circ)^2 - 4 \cdot 144$. Is it negative?
Typo: it should be $5^{2}=13^{2}+c^{2}-2\times 13c\cos 29{{}^\circ}$. –  Américo Tavares Feb 5 '12 at 20:49
@AméricoTavares Ah, thanks. Not so easy to calculate by hand now, although $\cos 30^\circ$ is quite close and gives one a sense of it. –  Dylan Moreland Feb 5 '12 at 20:57