Question 1023619: Hi, basically could somebody show me the steps as to how the word problem below was turned into the equation? I want to know how the quadratic equation was made step by step? The equation is at the bottom after the word problem. Many thanks
" TAHIR HAS A REGULAR CYCLE TRAINING ROUTE THAT IS 60KM LONG. IF TAHIR CYCLES 3KMH SLOWER THAN HIS RACING SPEED,IT TAKES HIM 3 HOURS AND 20 MINUTES LONGER TO COMPLETE THE ROUTE.
A) WORK OUT TAHIR'S RACING SPEED. STATE IN WORDS WHAT ALGEBRAIC METHOD YOU ARE USING AT EACH STAGE.
B) CHECK YOUR ANSWER BY USING TWO FURTHER ALTERNATIVE METHODS FOR SOLVING QUADRATIC EQUATIONS.
C) EVALUATE ALL THREE METHODS FOR EFFICIENCY AND EFFECTIVENESS.
V^2 - 3V - 54 = 0
TAHIR HAS A REGULAR CYCLE TRAINING ROUTE THAT IS 60KM LONG. IF TAHIR CYCLES 3KMH SLOWER THAN HIS RACING SPEED,IT TAKES HIM 3 HOURS AND 20 MINUTES LONGER TO COMPLETE THE ROUTE.
A) WORK OUT TAHIR'S RACING SPEED. STATE IN WORDS WHAT ALGEBRAIC METHOD YOU ARE USING AT EACH STAGE.
B) CHECK YOUR ANSWER BY USING TWO FURTHER ALTERNATIVE METHODS FOR SOLVING QUADRATIC EQUATIONS.
C) EVALUATE ALL THREE METHODS FOR EFFICIENCY AND EFFECTIVENESS."
V^2 - 3V - 54 = 0 (Please show me how the made that word problem into the equation? I know the answers are V = 9 or -6
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
if your final equation is v^2 - 9v - 54 = 0, then the factors are (v-9) * (v+6).
the basic equation is rate * time = distance.
distance equals 60 km.
speed = r kmph
time = t hours.
the basic equation becomes r * t = 60
when you are traveling 3 kmph slower, the rate becomes r-3.
when it takes 3 hours and 20 minutes longer, the time becomes t + 3 and 1/3.
your rate * time equation becomes (r-3) * (t + 3 + 1/3) = 60.
the distance is the same.
the rate is 3kmph slower.
the time is 3 and 1/3 hours longer.
you have 2 equations that need to be solved simultaneously.
they are r * t = 60 and (r-3) * (t + 3.33...) = 60
from r * t = 60, solve for t to get t = 60 / r.
in (r-3) * (t + 3.33...) = 60, replace t with 60 / r to get:
(r-3) * (60/r + 3.33...) = 60
multiply both sides of the equation by r to get:
(r-3) * (60 + 3.33... * r) = 60 * r
multiply the factors to get:
60r + 3.33... * r^2 - 180 - 10 * r = 60r
the 60r on the left side of the equation and the 60 r on the right side of the equation cancel out and you get:
3.33... * r^2 - 180 - 10 * r = 0
divide both sides of the equation by 3.33... (this is really 3 and 1/3), to get:
r^2 - 54 - 3r = 0
re-arrange the terms in descending order of degree to get:
r^2 - 3r - 54 = 0.
i used r for rate rather than v for velocity.
it doesn't matter.
the answer is the same.
from r^2 - 3r - 54, you can factor the equation to get:
(r-9) * (r+6) = 0
that makes r = 9 or -6.
r can't be negative, so r = 9.
when r = 9, r * t = 60 becomes 9 * t = 60.
solve for t to get t = 60/9 = 6 and 2/3.
9 * (6 + 2/3) = 60 so that part checks out.
your normal speed is 9 kmph and your normal time is 6 and 2/3 hours.
when you cut your speed by 3 kmph, then your speed is 6 kmph.
your time is increased by 3 and 1/3 hours.
that makes you time equal to 6 and 2/3 + 3 and 1/3 = 10 hours.
rate * time becomes 6 * 10 = 60 km.
answer checks out and so is good.
i'm not quite sure what you were asked to solve here.
if it was just to factor the quadratic, then there are 3 basic methods you can choose.
one is use of the quadratic formula.
two is the box method.
three is factor by grouping.
four is the guess and check method.
i don't have time to show you how to do them, but here's some references.
http://www.purplemath.com/modules/factquad.htm
http://www.purplemath.com/modules/factquad3.htm
http://www.regentsprep.org/regents/math/algtrig/atv1/revfactorgrouping.htm
http://www.regentsprep.org/regents/math/algtrig/atv1/revFactoring.htm
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut17_quad.htm
as you can see, there's lots of methods.
last one involves use of the quadratic formula.
http://www.purplemath.com/modules/quadform.htm
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