ExpandStr\( \mathrm { \displaystyle\frac{480}{v - 8} - \displaystyle\frac{480}{v} } \)ProcessLogRewrite \( \mathrm { \displaystyle\frac{480}{v - 8} - \displaystyle\frac{480}{v} } \) with using the common denominator \( \mathrm { \left(-8 + v\right) \cdot v } \) : \n \( \mathrm { \displaystyle\frac{480 \cdot v + 3840 - 480 \cdot v}{v \cdot \left(-8 + v\right)} } \)Str\( \mathrm { \displaystyle\frac{480 \cdot v + 3840 - 480 \cdot v}{v \cdot \left(-8 + v\right)} } \)ProcessLog\( \mathrm { 480 \cdot v - 480 \cdot v = 0 } \)Str\( \mathrm { \displaystyle\frac{0 + 3840}{v \cdot \left(-8 + v\right)} = 3 } \)Str\( \mathrm { \displaystyle\frac{3840}{v \cdot \left(-8 + v\right)} = 3 } \)StrProcessLogUse the distributive property of multiplication: \n \( \mathrm { a \cdot ( b + c ) = a \cdot b + a \cdot c } \) \n \( \mathrm { \displaystyle\frac{3840}{v \cdot \left(-8 + v\right)} = \displaystyle\frac{3840}{v^{2} - 8 \cdot v} } \)Str\( \mathrm { \displaystyle\frac{3840}{v^{2} - 8 \cdot v} = 3 } \)ProcessLogMultiply both sides with \( \mathrm { \displaystyle\frac{1}{3840} } \)Str\( \mathrm { \displaystyle\frac{3840}{3840} \cdot \displaystyle\frac{1}{v^{2} - 8 \cdot v} = 3 \cdot \displaystyle\frac{1}{3840} } \)ProcessLogSimplify the numerator and denominator: \n \( \mathrm { \displaystyle\frac{3840}{3840} \cdot \displaystyle\frac{1}{v^{2} - 8 \cdot v} = \displaystyle\frac{1}{v^{2} - 8 \cdot v} } \)ProcessLogSimplify the numerator and denominator: \n \( \mathrm { \displaystyle\frac{1}{1280 \cdot 3} \cdot 3 = \displaystyle\frac{1}{1280} } \)Str\( \mathrm { \displaystyle\frac{1}{v^{2} - 8 \cdot v} = \displaystyle\frac{1}{1280} } \)ProcessLogRaise both sides of the relation to the \( \mathrm { -1 } \)nd powerStr\( \mathrm { v^{2} - 8 \cdot v = 1280 } \)ExpandProcessLogDetermine the discriminant of the quadratic equation to find the roots : \n \( \mathrm { a x ^ { 2 } + b x + c \Longrightarrow \Delta = b ^ { 2 } - 4 a c } \) \n \( \mathrm { a = 1 } \) \n \( \mathrm { b = -8 } \) \n \( \mathrm { c = -1280 } \)Str\( \mathrm { \Delta = \left(-8\right)^{2} + \left(-4\right) \cdot \left(-1280\right) } \)ProcessLog\( \mathrm { \left(-4\right) \cdot \left(-1280\right) = 5120 } \)Str\( \mathrm { \Delta = \left(-8\right)^{2} + 5120 } \)ProcessLogSince discriminant greater than zero the equation has two distinct real roots. Use the quadratic formula \( \mathrm { x _{ 1, 2 } = \displaystyle\frac { - b \pm \sqrt { \Delta } } { 2 a } } \) to determine roots.StrFirst root: \( \mathrm { v_{1} = \displaystyle\frac{8 + \sqrt{\left(-8\right)^{2} + 5120}}{2} } \)ExpandProcessLogSimplify the root: \n \( \mathrm { \sqrt{\left(-8\right)^{2} + 5120} = 72 } \)Str\( \mathrm { v_{1} = \displaystyle\frac{8 + 72}{2} } \)ProcessLog\( \mathrm { 8 + 72 = 80 } \)Str\( \mathrm { v_{1} = \displaystyle\frac{80}{2} } \)ProcessLogSimplify the numerator and denominator: \n \( \mathrm { \displaystyle\frac{40 \cdot 2}{2} = 40 } \)Str\( \mathrm { v_{1} = 40 } \)\( \mathrm { v_{1} = \displaystyle\frac{8 + \sqrt{\left(-8\right)^{2} + 5120}}{2} } \)\( \mathrm { v_{1} = 40 } \)SimplifyStrSecond root: \( \mathrm { v_{2} = \displaystyle\frac{8 - \sqrt{\left(-8\right)^{2} + 5120}}{2} } \)ExpandProcessLogSimplify the root: \n \( \mathrm { \sqrt{\left(-8\right)^{2} + 5120} = 72 } \)Str\( \mathrm { v_{2} = \displaystyle\frac{8 - 72}{2} } \)ProcessLog\( \mathrm { 8 - 72 = -64 } \)Str\( \mathrm { v_{2} = - \displaystyle\frac{64}{2} } \)ProcessLogSimplify the numerator and denominator: \n \( \mathrm { \displaystyle\frac{\left(-32\right) \cdot 2}{2} = -32 } \)Str\( \mathrm { v_{2} = -32 } \)\( \mathrm { v_{2} = \displaystyle\frac{8 - \sqrt{\left(-8\right)^{2} + 5120}}{2} } \)\( \mathrm { v_{2} = -32 } \)Simplify\( \mathrm { v^{2} - 8 \cdot v - 1280 = 0 } \)\( \mathrm { v = \left\{-32, 40\right\} } \)Determine the roots\( \mathrm { \displaystyle\frac{480}{v - 8} - \displaystyle\frac{480}{v} = 3 } \)\( \mathrm { v = \left\{-32, 40\right\} } \)Isolate \( \mathrm { v } \)

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Solution: Train travels 480 km if speed had been 8 km/h less it takes 3 hours more

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