Question 1185065: Thrice the difference of the square of a positive number and 6 is equal to 3 less than for times the number. What is the number?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation is:
3 * (x^2 - 6) = 4x - 3
simplify to get:
3x^2 - 18 = 4x - 3
add 3 and subtract 4x from both sides of the equation, and order the terms in descending order of degree to get:
3x^ 2 - 4x - 15 = 0
factor that equation to get:
(3x + 5) * (x - 3) = 0
set each of the factors equal to 0 and solve for x to get:
3x + 5 = 0 gets you x = -5/3
x - 3 = 0 gets you x = 3.
since the number has to be positive, x = 3 must be your answer.
when x = 3:
3 * (x^2 - 6) = 3 * (9 - 6) = 3 * 3 = 9.
4 * x - 3 = 12 - 3 = 9
they're the same, confirming x = 3 is the answer.
there are various ways to factor the quadratic.
here's a reference:
https://www.purplemath.com/modules/solvquad6.htm
if all else fails, use the quadratic formula.
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