Answer
There are two solutions for $\alpha$:
${\alpha _1} = \frac{1}{{13}}\left( {16 + \sqrt {347} } \right)$ and ${\alpha _2} = \frac{1}{{13}}\left( {16 - \sqrt {347} } \right)$
Work Step by Step
We have ${\bf{v}} = \left( { - 2,5} \right)$, ${\bf{w}} = \left( {3, - 2} \right)$ and $||{\bf{v}} + \alpha {\bf{w}}|| = 6$.
Since $||{\bf{v}} + \alpha {\bf{w}}|{|^2} = \left( {{\bf{v}} + \alpha {\bf{w}}} \right)\cdot\left( {{\bf{v}} + \alpha {\bf{w}}} \right),$, so
$||{\bf{v}} + \alpha {\bf{w}}|{|^2} = {\bf{v}}\cdot{\bf{v}} + \alpha {\bf{w}}\cdot{\bf{v}} + \alpha {\bf{v}}\cdot{\bf{w}} + {\alpha ^2}{\bf{w}}\cdot{\bf{w}}$
$||{\bf{v}} + \alpha {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2\alpha {\bf{v}}\cdot{\bf{w}} + {\alpha ^2}||{\bf{w}}|{|^2}$
Evaluate $||{\bf{v}}|{|^2}$, ${\bf{v}}\cdot{\bf{w}}$ and $||{\bf{w}}|{|^2}$:
$||{\bf{v}}|{|^2} = {\left( { - 2} \right)^2} + {5^2} = 29$
${\bf{v}}\cdot{\bf{w}} = \left( { - 2,5} \right)\cdot\left( {3, - 2} \right) = - 16$
$||{\bf{w}}|{|^2} = {3^2} + {\left( { - 2} \right)^2} = 13$
So,
$||{\bf{v}} + \alpha {\bf{w}}|{|^2} = 29 - 32\alpha + 13{\alpha ^2}$
Since $||{\bf{v}} + \alpha {\bf{w}}|| = 6$,
$29 - 32\alpha + 13{\alpha ^2} = 36$
$13{\alpha ^2} - 32\alpha - 7 = 0$
The solutions to this quadratic equation:
$\alpha = \frac{{32 \pm \sqrt {{{\left( { - 32} \right)}^2} - 4\cdot13\cdot\left( { - 7} \right)} }}{{2\cdot13}} = \frac{1}{{13}}\left( {16 \pm \sqrt {347} } \right)$
So, there are two solutions for $\alpha$:
${\alpha _1} = \frac{1}{{13}}\left( {16 + \sqrt {347} } \right)$ and ${\alpha _2} = \frac{1}{{13}}\left( {16 - \sqrt {347} } \right)$
