Susan C. answered • 01/26/18
Tutor
5
(31)
I love math, and I love to teach it.
Hello, T.
The above equation is a quadratic equation when X is to the second power. Quadratics on the graph usually cross the x-axis in two places unless the quadratic has a double root or no real solutions.
They are giving you a restriction, x>0, which as you know, means that the value of "x" must be positive and x≠0 but greater than it. Since this is a difficult equation to factor easily, I suggest you use the Quadratic Formula, which is
x= [-b +/- √(b2-4ac) ] 3x2-5x- (11/12) =0 a=3 b= -5 c= -11/12
2a
1. Substitute the values for "a, b," and "c" into the Quadratic Equation.
2. x= - (-5) +/- √( ( -5)2 - 4(3)( -11/12) ) = 5 +/- √( (25)+ 12(11/12) )
2(3) 6
3. x= 5+/- √(25+11) The +/- means that there are two answers: x= 5 +√(36) = (5+6)/6=11/6≅1.83
6 6
and x= (5-√36)= (5-6)/6 =-1/6 ≅ -.17 A negative answer is less than zero.
6
The negative answer does not fit the restriction, so don't use it in the choices. Does any choice match the other answer of x≅ 1.83? Let's look at each possible choice. A) 0<x<1 a decimal less than 1 but greater than 0
No, 1.83 is greater than 1, not less than it.
b) 1<x<2 "x" is greater than one but less than 2. Is 1.83 greater than one but less than 2? Yes!
As for choices C and D, they are both false as related to the value 1.83.
I hope that this example helps. Susan C.
