Bunuel wrote:
5. What is the value of integer x?(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.
(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.
Answer: D.
Check this for more on solving inequalities like the one in the first statement:
x2-4x-94661.html#p731476 inequalities-trick-91482.htmleverything-is-less-than-zero-108884.html?hilit=extreme#p868863xy-plane-71492.html?hilit=solving%20quadratic#p841486Hope it helps.
Hey Bunuel,
Two questions.
1) How did you factor 2x^2+9<9x (ie 2x^2-9x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method.
2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"??
I used the dumb method of just plugging values that lie from (-infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)?
Many thanks Bunuel! Your my hero dude!