Question

P is directly proportional to (q + p)^2. Q is 1 when p is 1. Find the equation connectng q to p and find q when p is 10

Answer 1

A directly proportionality between x and y is written as:

y = k*x

where k is the constant of proportionality.

Then, if p is directly proportional to (q + p)^2

We can write this as:

p = k*(q + p)^2

Now we know that q = 1 when p = 1

Then we can replace these values in the above equation to find the value of k

1 = k*(1 + 1)^2

1 = k*2^2 = k*4

1/4 = k

Then the equation is:

p = (1/4)(p + q)^2

Now we want to find the value of q when p = 10

So we just replace p by 10, and solve the equation

10 = (1/4)*(q + 10)^2

10*4 = (q + 10)^2

40 = q^2 + 2*q*10 + 10^2

40 = q^2 + 20*q + 100

This is a quadratic equation for q, such that:

q^2 + 20*q + 100 - 40 = 0

q^2 +20*q + 60 = 0

This is the equation that gives the value of q when p = 10

To solve this, we can use Bhaskara's equation, which says that for a quadratic equation:

a*x^2 + b*x + c = 0

The solutions are given by:

`x = (-b \pm √(b^2 - 4*a*c) )/(2*a)`

In this case, the solutions for q are:

`q = (-20 \pm √(20^2 - 4*1*60) )/(2*1) = (-20 \pm 12.65)/(2)`

Then we have two possible values for q:

q = (-20 + 12.65)/2 = -3.675

q = (-20 - 12.65)/2 = -16.325