# Quadratic equations

**Question**

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.

**Answer**

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. 🙂