Speed of car

The car went to a city that was 240 km away. If his speed increased by 8 km/h, it would reach the finish one hour earlier. Determine its original speed.

Correct result:

v =  40 km/h

Solution:

s=vt  240=vt 240=(v+8)(t1)  240=(v+8)(240/v1)  240v=(v+8)(240v)  240 v=(v+8) (240v) v2+8v1920=0  a=1;b=8;c=1920 D=b24ac=8241(1920)=7744 D>0  v1,2=b±D2a=8±77442 v1,2=8±882 v1,2=4±44 v1=40 v2=48   Factored form of the equation:  (v40)(v+48)=0  v>0  v=v1=40 km/hs=vt \ \\ \ \\ 240=vt \ \\ 240=(v+8)(t-1) \ \\ \ \\ 240=(v+8)(240/v-1) \ \\ \ \\ 240*v=(v+8)*(240-v) \ \\ \ \\ 240 \cdot \ v=(v+8) \cdot \ (240-v) \ \\ v^2 +8v -1920=0 \ \\ \ \\ a=1; b=8; c=-1920 \ \\ D=b^2 - 4ac=8^2 - 4\cdot 1 \cdot (-1920)=7744 \ \\ D>0 \ \\ \ \\ v_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -8 \pm \sqrt{ 7744 } }{ 2 } \ \\ v_{1,2}=\dfrac{ -8 \pm 88 }{ 2 } \ \\ v_{1,2}=-4 \pm 44 \ \\ v_{1}=40 \ \\ v_{2}=-48 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (v -40) (v +48)=0 \ \\ \ \\ v>0 \ \\ \ \\ v=v_{1}=40 \ \text{km/h}

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