Answer:
[tex]\large \boxed{\sf \ \ \ \dfrac{\sqrt{4025}-5}{2}=29.22144... \ \ \ }[/tex]
Step-by-step explanation:
Hello
Let's note the original speed of the car v
it means that in 1 hour he is going v miles
so to go 200 miles it takes ( in hour)
[tex]\dfrac{200}{v}[/tex]
If the speed of the car is v+5 than to go 200 miles it takes (in hour)
[tex]\dfrac{200}{v+5}[/tex]
and this time is one hour less so we can write
[tex]\boxed{\sf \ \ \dfrac{200}{v+5}=\dfrac{200}{v}-1 \ \ }[/tex]
We can multiply by v(v+5) both parts of the equation so
[tex]200v=200(v+5)-v(v+5)\\\\<=>200v=200v+1000-v^2-5v\\\\<=>v^2+5v-1000=0[/tex]
[tex]\Delta=b^2-4ac=5^2+4*1000=4025\\\\ \text{There are potential solutions }\\\\\ \ \ \ \ x_1=\dfrac{-5-\sqrt{4025}}{2}\\\\\ \ \ \ \ x_2=\dfrac{-5+\sqrt{4025}}{2}[/tex]
Only one is positive and this is is
[tex]x_1=\dfrac{\sqrt{4025}-5}{2}=29.22144...[/tex]
So the original speed is 29.22144... mph
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
