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Can someone please help me with these math Questions..10 pts to correct answer! (please show work)<br /><br />1.) A student invests $8,500 in a savings account drawing interest that is compounded annually. Find the annual rate if the money grows to $9,193.60 in 2 years<br /><br />2.) The sum of the squares of two consecutive postive integers is 85. Find the integers.<br /><br />3.) Find a third-degree equation that has a solution set of <br />{2,3,-4}<br />

vicki

1)
x = Interest rate
End of first year = y = 8500 + 8500x
End of second year = y + (y x) = 9193.60

Now we substitute y
8500 + 8500 x + ((8500 + 8500x)x) = 9193.60
8500 + 8500 x + 8500x + 8500x^2 = 9193.60
8500 + 17000 x + 8500x^2 = 9193.60
8500x^2 + 17000 x + 8500 = 9193.60
8500 (x^2 + 2x + 1) = 9193.60
x^2 + 2x + 1 = 9193.60 / 8500
x^2 + 2x + 1 = 1.0816
x^2 + 2x + 1 - 1.0816 = 0
x^2 + 2x - 0.0816 = 0

Using quadratic equation
x = -b +- sqrt(b^2 - 4ac) / 2a
where a = 1, b = 2, c = -0.0816

x = -2 +- sqrt(2^2 - 4(1)(- 0.0816)) / 2(1)
x = -2 +- sqrt(4 + 0.3264) / 2
x = -2 +- sqrt(4.3264) / 2
x = -2 +- 2.08 / 2
x = 0.08 / 2
x = 0.04

We disregard the negative number resulted from the quadratic equation since the interest rate is a positive number

So the interest rate was 4%

2)
x^2 + (x + 1)^2 = 85
x^2 + x^2 + 1 + 2x = 85
2x^2 + 2x + 1 = 85
2x^2 + 2x + 1 - 85 = 0
2x^2 + 2x - 84 = 0
2(x^2 + x - 42) = 0
x^2 + x - 42 = 0

Using quadratic equation
x = -b +- sqrt(b^2 - 4ac) / 2a
where a = 1, b = 1, c = -42

x = -1 +- sqrt(1^2 - 4(1)(- 42)) / 2(1)
x = -1 +- sqrt(1 + 168) / 2
x = -1 +- sqrt(169) / 2
x = -1 +- 13 / 2
x = 12 / 2
x = 6

We disregard the negative number resulted from the quadratic equation since question says the squares of two consecutive positive integers

So the numbers are 6 and 7

3.
(x - 2)(x - 3)(x + 4) = 0
(x^2 - 2x - 3x + 6)(x + 4) = 0
(x^2 - 5x + 6)(x + 4) = 0
(x^3 - 5x^2 + 6x + 4x^2 - 20x + 24) = 0
x^3 - x^2 - 16x + 24 = 0

I hope this helped

Kia