What is the integer if twice the square of a positive integer is 35 more than 9 times that integer?

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2006-12-03 06:45:04 · 3 answers · asked by brandy 1 in Science & Mathematics ➔ Mathematics

2x² -35 = 9x
2x² - 9x -35 = 0
(2x + 5)(x - 7)=0

The problem says that you're looking for an integer, so x=7.

2006-12-03 06:50:03 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋

you first have to make an equation based on the information
Let x = integer
2x^2=9x + 35
2x^2 - 9x - 35 = 0
Then, solve the quadratic to find x

2006-12-03 06:49:42 · answer #2 · answered by Panky1414 2 · 0⤊ 0⤋

2x^2 = 35 + 9x
2x^2 - 9x - 35 = 0
(2x+5)(x-7) = 0
x = -2.5 or 7.

So x = 7.

2006-12-03 06:48:24 · answer #3 · answered by stephen m 4 · 0⤊ 0⤋