Question

If a firm invests amount P0 at an interest rate of​ r%, then a year later invests amount P1 at the same​ rate, then the total amount at the end of two years is given by the formula = P0​(1+​r)^0.5 + P1​(1+​r). A firm invests ​$1000​, then a year later invests an additional ​$2400. At the end of two​ years, it has a total of ​$3623. What is the interest​ rate?

Answer 1

Answer:

interest rate = 5.01%

Explanation:

$a (1+r)^{2} + b (1+r) = c$

p0= a = 1000

p1= b = 2400

amount = c= -3623

rate = ?

Because the first amount is investment for a period of 2 years, and the second 1 year, we can solve for rate using the quadratic equation:

$x = \frac{ - b +/- \sqrt{b^{2} - 4ac} }{2a}$

$1000 (1+r)^{2} + 2400 (1+r) = 3623$

$1000 (1+r)^{2} + 2400(1+r) - 3,623 = 0$

A = 1000

B = 2400

C = -3623

$x = \frac{ - 2400 \sqrt{2400^{2} - 4*1000*-3623} }{2*1000}$

x1 = 1.0501111083677623

x2 = -3.4501111083677625

We use the positive root:

x1 = 1.0501111083677623 = (1+r)

1.0501111083677623 - 1 = r = 0.0501111 = 5.01%

EDIT several problems with the math tool but kind of worked