M27-14

Yes, \(2x^2 - 9x +9<0\) can be factored as \((x - 3)(2x - 3)<0\) but notice that we can factor 2 out of (2x - 3) and reduce by it to get \((x-3)(x-\frac{3}{2}) \lt 0\).

Theory on Inequalities
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Bunuel
the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3.
Can you please elaborate on this statement ? How did we reach that conclusion?

Thank you

Check here: solving-quadratic-inequalities-graphic-approach-170528.html

Bunuel Can you please explain statement 'B'? The part I missed was that why didn't you consider the negative of the LHS, and just the +ve?

Absolute value on the left hand side (|x + 10|) cannot be negative: \(|some \ expression|\geq{0}\). Thus 2x + 8, on the right hand side must also be non-negative, which means that \(x \ge -4\). Now, if \(x \ge -4\), then x + 10 will be positive, and we know that if x>0, then |x| = x, so |x + 10| = x + 10.

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Hope this helps.

Can you please tell, why are we taking Absolute value as positive?
I have seen in majority of the questions that when we solve Absolute value equation, it gives two solutions. one negative and other positive
For Example: the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html

can you please explain the difference? where we have take both values and where we have to take only positive value

The absolute value of a number is the value of a number without regard to its sign.

For example, |3| = 3; |-12| = 12; |-1.3|=1.3...

Another way to understand absolute value is as the distance from zero. For example, |x| is the distance between x and 0 on a number line.

From that comes the most important property of an absolute value: since the distance cannot be negative, an absolute value expression is ALWAYS more than or equal to zero.

Hi Bunuel,

Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.

Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...

Yes I understand that (x-3) (2x-3)<0 ... I then did the following:

x-3<0 ... so x<3
2x-3<0 ... so 2x<3...x<1.5

How can x be less than 3 and less than 1.5 and thus be an integer 2?

I clearly went wrong somewhere but I don't know where.

Thanks,

Tosin

(x - 3)(2x - 3) < 0 to be true the multiples, x - 3 and 2x - 3 must have different signs.

x - 3 > 0 and 2x - 3 < 0 --> x > 3 and x < 3/2. Both cannot be simultaneously true. So, this case is out.

x - 3 < 0 and 2x - 3 > 0 --> x < 3 and x > 3/2. So, 3/2 < x < 3.

Thanks Bunuel. Could you quickly confirm my understanding:
1) Test positive case for |x+10|, i.e. x+10 = 2x+8, i.e. x = 2
2) Test negative case for |x+10|, i.e. -x-10 = 2x+8, i.e. x = -6 => not a solution because of constraint x >= -4
Thus, x = 2 as only solution, right?

If you figured that x must be more than or equal to -4, then you could directly get that |x + 10| = x + 10 because if x >= -4, then x + 10 is positive, thus |x + 10| = x + 10 (recall that |a| = a if a >= 0).

If you want to open modulus in conventional way then you should consider two cases: x <= -10 and x >= 10.

Hi Bunuel
Thanks for your explanations
I have a real problem with factorizing quadratics containing "a>1 or a<1" coefficient (ax2+bx+c).
To me it's a real headache & nightmare, making me sometimes to use formula to find the roots which is absolutely time consuming and under the pressure of the test, mostly doesn't work at all.
I would be grateful if you share me anything on "how to factorize a quadratic equation with a non-1 coefficient for x2"

Factoring Quadratics
Solving Quadratic Equations

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For more check Ultimate GMAT Quantitative Megathread

Hope it helps.

Bunuel

For Statement B : |x+10|=2x+8 ,we can consider x+10= 2x+8 and x+10 = - [2x+8] and this gives 2 values 2 and -6 and so we can say B is not sufficient. I also understand your and other justification how B can be correct as well but why I am posting question is because I have seen other problems and solution on our club where when given mod of X , we consider both values of X while doing calculations. So want to understand when to consider both values + and - and when to consider only + value ?

When you solve the way you did, you should plug back the roots to confirm that they do in fact satisfy the equation. If you plug x = -6 you'd get that LHS = 4 and RHS = -4. 4 ≠ -4. So, x = -6 does not satisfy |x + 10| = 2x + 8.


Or if you adopt more conventional way you'd have:

If x < -10, then x + 10 < 0, so |x + 10| = -(x + 10) and in this case we'd have -(x + 10) = 2x + 8 --> x = -6. You should discard this solution because x = -6 is not in the range we are considering (x < -10).

If x >= -10, then x + 10 >= 0, so |x + 10| = x + 10 and in this case we'd have x + 10 = 2x + 8 --> x = 2. This solution is valid because x = 2 is in the range we are considering (x >= -10).

So, |x + 10| = 2x + 8 has only one solution: x = 2.

Hope it's clear.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

When you solve the way you did, you should plug back the roots to confirm that they do in fact satisfy the equation. If you plug x = -6 you'd get that LHS = 4 and RHS = -4. 4 ≠ -4. So, x = -6 does not satisfy |x + 10| = 2x + 8.

I also solved statement 2 the conventional way and arrived at X=2 or X=-6. I didn't understand your explanation why this statement would be sufficient until you said:'' you should plug back those values for X, which are 2 or -6, to see if LHD and the RHS of the equation would still be equal''. And yes only when X=2 are the LHS and the RHS equal.

Sometimes are the very tiny details that make the difference in understanding a question.

Thank you Bunuel!