Hi.
I thought we cannot square both sides (|12x−5|>|7−6x|) unless we know that they are positive.
Thanks!
Square both sides: 144x^2-120 x+25>36 x^2-84 x+49 --> 9 x^2-3x-2>0 --> factor: (x+\frac{1}{3})(x-\frac{2}{3})>0 (check here: http://www.purplemath.com/modules/factquad.htm). ">" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus x<-\frac{1}{3} OR x>\frac{2}{3}.
x=-4<-\frac{1}{3} and x=3>\frac{2}{3} are possible values of x --> the product = -12;
x=-\frac{7}{5}<-\frac{1}{3} and x=1>\frac{2}{3} are possible values of x --> the product = -7/5;
x=-\frac{4}{9}<-\frac{1}{3} and x=-1<-\frac{1}{3} are possible values of x --> the product = 4/9;
x=-1<-\frac{1}{3} and x=17>\frac{2}{3} are possible values of x --> the product = 17.
Answer: C.
I thought we cannot square both sides (|12x−5|>|7−6x|) unless we know that they are positive.
Thanks!
Bunuel wrote:
emmak wrote:
If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?
A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17
A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17
Square both sides: 144x^2-120 x+25>36 x^2-84 x+49 --> 9 x^2-3x-2>0 --> factor: (x+\frac{1}{3})(x-\frac{2}{3})>0 (check here: http://www.purplemath.com/modules/factquad.htm). ">" sign indicates that the solutions lies to the left of the smaller root and to the right of the greater root (check here: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus x<-\frac{1}{3} OR x>\frac{2}{3}.
x=-4<-\frac{1}{3} and x=3>\frac{2}{3} are possible values of x --> the product = -12;
x=-\frac{7}{5}<-\frac{1}{3} and x=1>\frac{2}{3} are possible values of x --> the product = -7/5;
x=-\frac{4}{9}<-\frac{1}{3} and x=-1<-\frac{1}{3} are possible values of x --> the product = 4/9;
x=-1<-\frac{1}{3} and x=17>\frac{2}{3} are possible values of x --> the product = 17.
Answer: C.