b44108…: “Opposite a fall calculation”

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Consider the vector $\vec {u} = \vector {a,a-1}$ . For how many values of $a$ does the vector $\vec {u} = \vector {a,a-1}$ have magnitude $\sqrt {5}$ ?

$0$ $1$ $2$ $3$ more than three, but finitely many infinitely many

The values of $a$ for which $|\vec {u}| = \sqrt {5}$ are $a= \answer {-1}$ and $a=\answer {2}$ . (type the smaller of the values first)

The magnitude of $\vec {u}$ in terms of $a$ is:

$|\vec {u}| = \sqrt {\answer {a^2+(a-1)^2} }$

Setting $|\vec {u}| = \sqrt {5}$ and squaring both sides gives:

$\answer {5} = \answer {a^2+(a-1)^2}$

After a little algebra, this gives a quadratic equation:

$\answer {2}a^2 +\left (\answer {-2}\right )a+\left (\answer {-4}\right ) = 0$

This can be factored, and the roots are $a=\answer {-1}$ and $a=\answer {2}$ . (type the smaller root first)