Question

Another of Bhaskara's problems results in a quadratic equation Parthava was enraged and seized a certain number of arrows to slay Karna. He expended one-half of them in defending himself. Four times the square root of the number of arrows were discharged against the horses. With six more, he transfixed Shalya, the charioteer. With three more, he rent the parasol, the standard, and the bow; and with the last one he pierced the head of Karna. How many arrows did Parthava have?

Answer 1

Parthava had 100 arrows.

Explanation:

Let's define N as the number of arrows that Parthava originally has.

He uses one-half of them in defending himself, so he used N/2 arrows

Now he uses four times the square root of the number of arrows, so now he uses:

4*√N

Then he uses 6

Then he uses 3

Then he uses the last one.

If we add all these numbers of arrows that he used, we should get the initial number of arrows that he used, then:

N/2 + 4*√N + 6 + 3 + 1 = N

Now we have an equation that we can try to solve.

First, let's move all the terms to the same side:

N/2 + 4*√N + 6 + 3 + 1 - N = 0

now we can simpify it:

(N/2 - N) + 4*√N + (6 + 3 + 1) = 0

-(1/2)*N + 4*√N + 10 = 0

Now we can define a new variable x = √N

Then we have: x^2 = N

now we can replace these new variables in our equation to get:

-(1/2)*x^2 + 4*x + 10 = 0

Now we just have a quadratic equation.

Remember that for a quadratic equation of the form:

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

The solutions were given by:

`x = (-b \pm √(b^2 - 4*a*c) )/(2a)`

Then in our case, the solutions will be:

`x = (-4 \pm √(4^2 - 4*(-1/2)*10) )/(2*(-1/2)) = (-4 \pm 6 )/(-1) = 4 \pm 6`

So there are two solutions:

x = 4 + 6 = 10

x = 4 - 6 = -2

And remember that x = √N

Then x should be positive, then we take x = 10 as our solution here.

then we can use the equation:

x = 10 = √N

then

10^2 = √N^2 = N

10^2 = 100 = N

Parthava had 100 arrows.