If an amount of money P in pesos is invested at r percent compounded annually, it will grow to an amount A=P(1+r)² in two years. Suppose Mrs. Madrigal wants her money amounting to Php 200 000 to grow to Php 228 980 in two years. At what rate must she invest her money?


I need the process of the solution.. please help me.. thanks


Sagot :

We have A=228,980 and P=200,000

A = P(1+r)²
228,980 = 200,000(1+r)²
228,980 = 200,000(r²+2r+1)
228,980 = 200,000r² + 400,000r + 200,000
0 = 200,000r² + 400,000r - 28,980
0 = 10,000 r² + 20,000r - 1,449

We then use the quadratic formula:
r = -20,000 ± √(20,000²+4·10,000·1449) = -20,000 ± √457960000 
                     2(10,000)                                    20,000)
= -20,000 ± 21,400  = -100 ± 107 = -207/100 or 7/100
          20,000                 100

We take the positive rate which is 7%. So r=7%.
Hmm, now I can post the solution here I have posted in the comment section ten hours ago (check above).

A= 228, 980
P= 200, 000
r = rate 

228,980 = 200,000 (1 + r)²
228,980 = 200 000 (1 + 2r + r²)
228,980 = 200 000 + 400 000 r + 200 000r²
200 000 r² + 400 000 r + 200 000 - 228 980 = 0
2000 000 r² + 400 000 r -28 980 = 0

Factor or use the quadratic formula:

( x - 7/100 ) (x + 207/100) = 0

x = 7/100    or x = -207/100

Choose the positive root, x-value: 7/100

7/100 = 0.07 = 7 percent

The rate is 7 %