Quadratic equations... 10 pts for best answer!?

Question

ok, i dont normally do this but its quite urgent, i have maths homework and im running late plz help!

1) A number exceeds four times its reciprocal by 3. find the number forming a quadratic equation.

2) A cyclist travels 40km at a speed of x km/h. Find the time taken in terms of x. Find when his speed is reduced by 2 km/h. If the difference between the times is one hour, find the original speed x forming a quadratic equation.

Plz if u can help... help me :P i know i should be doing my own homework but i have like 30 questions like these still to do! If u can answer one of them, answer... n if both then YAY! i'll give 10 pts for the best answer that will also explain the woking out. THANKS!

Answer 1

1) Number = X and Reiprocal = 1 / X.
four times its reciprocal = 4* ( 1/X) = 4/X.
"Number exceeds 4 times reiprocal by 3" means the "number = X" is equal to "4/X + 3"
Information says : X = 4 / X + 3. Multiply by X to get the quadratic below,
X^2 = 4 + 3X or,
X^2 - 3x - 4 = 0
(X - 4) ( X + 1 ) = 0
So X = 4 or X = -1 is the answer.

2) Speed reduced by 2km/h = x - 2.

Remember that from Speed = distance / time, we get,
time = distance / speed

Original time = time taken to travel 40 km with a speed of x
= 40 / x = "shorter time because speed is faster"

New Time = time taken to travel 40km with speed of (x-2)
= 40 / (x - 2) = "Longer time becase speed is lower"

Difference between the times = 1. So,

"longer time" minus "shorter time = 1.
40 / (x - 2) - 40/x = 1

40x - 40*(x - 2) = x*(x - 2)
40x - 40x + 80 = x^2 - 2x
x^2 - 2x - 80 = 0
(x - 10)(x + 8) = 0
So,
x = 10 or x = - 8. Since "x" here represents "speed" its value is positive, i.e. >0. So x = 10 km/h or 10 kilometer per hour.

Only "velocity" has "positive" and "negative" values depending on directions. :)

Hope this helped.

:)

Answer 2

1) let the number be x

reciprocal of this number 1/x
x exceeds 4times 1/x by3
x= 4/x+3
x^2= 4+3x
x^2- 3x-4=0
x^2- 4x+x-4=0
x(x-4)+1(x-4)=0
(x+1)(x-4)=0
x =4,-1

2)speed = distance/time
speed * time = distance

let the time taken to travel a 40km be x
when speed is 2 km/h difference between time is 1hour
2*x =40
x=40/2
x=20
so the time taken to travel 40 km at a speed of 2 km/hr is 20 hrs

the differernce in time is 1 hour, so the time is 20hr -1 hr
which is 19 hrs
so the same distance of 40 km in 19hrs
speed is = 40/19
speed = 2.105 km/hr

Answer 3

(1) x-4/x =3, x^2-4=3x, x^2-3x-4=0
factorising (x-4)(x+1)=0 therefore x = 4 and -1

(2) speed = distance/time
x=40/t and t=40/x

if time taken at x km/h is t1 and time taken at(x-2) km/h is t2
then t2-t1 = 1
substituting
40/(x-2) - 40/x =1
multiplying by (x^2-2x)
40x - 40(x-2) = x^2-2x
re-arranging
x^2-2x-80=0
factorising
(x-10)(x+8) = 0
and x=10 km/h

Answer 4

you need to use the quadratic formulation, yet for that reason it relatively is factorable. 5x^2 + 11x + 6 = 0 All words are beneficial, so it relatively is in the style (x+a)(x+b). the climate of five are 5 and a million. the climate of 6 are a million, 2, 3, and six. With slightly trial and blunder (in the beginning, after prepare you will see the respond in the present day), you spot that (5x+6)(x+a million) = 0. utilising the 0 product sources (if xy = 0, the x and/or y ought to be 0), we see that x+a million=0 and 5x+6 =0. With slightly common algebra, we see that x = -a million and -6/5

Answer 5

let x be the number

x - 4/x = 3
x^2 - 4 = 3x
x^2 - 3x - 4 = 0
(x-4)(x+1) = 0
x = 4 or -1
check
4 - 4/4 = 3 and -1 - 4/(-1) = 3


the time it takes = distance / speed = 40/x
the equation is

40/(x-2) - 40/x = 1
40x - 40(x-2) = x(x-2)
x^ - 2x - 80 = 0

Answer 6

1,
x + 3 = 4(1/x)
x = 4/x -3
x² = 4 - 3x
x² + 3x = 4
ANS: x = 1

2,
time = distance/speed
ANS#1: time = 40/x
ANS#2: time = 40/(x-2)
40/(x -2) -40/x = 1
40x -40x +80 = x² -2x
x² - 2x - 80 = 0
(x -10)(x +8) = 0
ANS#3: x = 10km/h