Problem Solving

Hello experts,

What is the range of solutions for |x^2-4| > 3x ?

I solved it like this: https://s25.postimg.cc/a9mppfx0v/Graph.png and evidently, it yields an incorrect solution. What am I missing here?

As it is, inequalities get on me nerves and add to that an absolute operator. Gee... does GMAT ever get any easier on people?

[email protected] wrote:Hi Blackboxx,

What is the source of this question? Did it come from a GMAT book or from a math book? And where are the answer choices? Having those answer choices would give you some valuable "hints" as to what the solution would be.

Hello Rich - Thanks for the super quick response. Sadly, I cannot remember where I saw this question and therefore cannot cite the source (I posted the same question on a couple of other forums though). If not citing a source means breaking this forum's rules, then I am truly sorry for my ignorant mistake. Mods can kindly take my question down.

|x^2-4| > 3x

With the given inequality, there are a couple of things that immediately come to mind regarding possible values of X (TESTing VALUES and using some basic Number Properties):

1) X can = 0
2) ANY negative value of X will "fit" the inequality
3) X CANNOT = 1

With these 3 deductions, we might have enough information to answer the question (if it's a GMAT question with answer choices to work with) and any other work might not be necessary.

The answer is x<1, x>4. How can this question be answered using the "plot on the number line" approach? Since, I tend to use that approach for all the inequalities question, I'd like to see how it applies to this question too.

TIA

Deduction aside, how can I plot the solution on the number line?

Blackboxx wrote: What is the range of solutions for |x^2-4| > 3x ?

Determine the CRITICAL POINTS.
The critical points are the values of x where the left-hand side is equal to the right-hand side:
|x² - 4| = 3x.
Since an absolute value cannot be negative, both sides of the equation above must be NONNEGATIVE, implying that x≥0.
Further, the solutions for x are almost certain to be INTEGER VALUES.
(On the GMAT, the roots of a quadratic are almost always integer values.)

Test nonnegative integer values in |x² - 4| = 3x:
x=0 --> |0² - 4| = 3*0 --> 4=3
x=1 --> |1² - 4| = 3*1 --> 3=3
x=2 --> |2² - 4| = 3*2 --> 0=6
x=3 --> |3² - 4| = 3*3 --> 5=9
x=4 --> |4² - 4| = 3*4 --> 12=12

No values greater than 4 are viable.
If x increases beyond 4, the value of |x² - 4| will become too large to equal 3x.

Thus, only the options in red yield valid solutions for x, implying two critical points:
x=1, x=4.
These are the only values of x where |x² - 4| = 3x.
To determine where |x² - 4| > 3x, test one value to the left and right of each critical point.

x<1:
Plugging x=0 into |x² - 4| > 3x, we get:
|0² - 4| > 3*0
4 > 0.
This works.
Thus, x<1 is a valid range.

1<x<4:
Plugging x=2 into |x² - 4| > 3x, we get:
|2² - 4| > 3*2
0 > 6.
Doesn't work.
Thus, 1<x<4 is NOT a valid range.

x>4:
Plugging x=5 into |x² - 4| > 3x, we get:
|5² - 4| > 3*5
21 > 15.
This works.
Thus, x>4 is a valid range.

Thus, the valid ranges for |x² - 4| > 3x are x<1 and x>4.
Plotted on a number line:
<-----1.............4----->

Algebra:

The critical points for |x² - 4| = 3x can also be determined algebraically.
As noted previously:
Since an absolute value cannot be negative, both sides of the equation above must be NONNEGATIVE, implying that x≥0.
Thus, only NONNEGATIVE solutions for x are valid.

Case 1: Signs unchanged
x² - 4 = 3x
x² - 3x - 4 = 0
(x-4)(x+1) = 0.
x=4 or x=-1.
Since x must be nonnegative, Case 1 has only one valid solution:
x=4.

Case 2: Signs changed on ONE SIDE
x² - 4 = -3x
x² + 3x - 4 = 0
(x+4)(x-1) = 0.
x=-4 or x=1.
Since x must be nonnegative, Case 2 has only one valid solution:
x=1.

Thus, the critical points are x=1 and x=4.
These are the only values of x where |x² - 4| = 3x.
To determine where |x² - 4| > 3x, test one value to the left and right of each critical point, as shown in my initial solution above.

Many range problems can also be solved by PLUGGING IN THE ANSWERS.
One example:
https://www.beatthegmat.com/absolute-val ... 74256.html

More practice with critical points:
https://www.beatthegmat.com/if-x-is-not- ... 65585.html
https://www.beatthegmat.com/inequality-c ... 89518.html
https://www.beatthegmat.com/knewton-q-t89317.html
https://www.beatthegmat.com/which-is-true-t89111.html
https://www.beatthegmat.com/value-of-x-t268225.html
https://www.beatthegmat.com/if-n-is-not- ... 76305.html

GMATGuru - I love how amazingly simple your Control Points method is. Does it work for any inequality? I have cross checked on a couple of expressions and it DID! But what if there is a <= or >=, should I just replace it with < and > in the final range?

Finally, I have a general question around absolute expressions. For e.g., the way I used to find ranges until I saw Rich's and your explanations, was to directly solve the expression.

Given,

|x^2-4| > 3x

I would bifurcate this expression as (x^2-4) and -(x^2-4). Now my question specifically is when I pick the negative value. Should I flip the direction and multiply the RHS? In other words, does the expression become ...

-(x^2-4) < - 3x

Please confirm! Thanks.

Best,
Jay

Blackboxx wrote:But what if there is a <= or >=, should I just replace it with < and > in the final range?

|x²-4| > 3x.
Here, the left-hand side is NOT allowed to equal the right-hand side.
As shown above, the valid ranges for this inequality are x<1 and x>4.

|x²-4| ≥ 3x.
Here, the left-hand side IS allowed to equal the right-hand side.
Thus, the critical points -- the values of x that make the two sides equal -- are VALID solutions and must be INCLUDED in the solution set.
Result:
The ranges where |x²-4| ≥ 3x are x≤1 and x≥4.

Finally, I have a general question around absolute expressions. For e.g., the way I used to find ranges until I saw Rich's and your explanations, was to directly solve the expression.

Given,

|x^2-4| > 3x

I would bifurcate this expression as (x^2-4) and -(x^2-4). Now my question specifically is when I pick the negative value. Should I flip the direction and multiply the RHS? In other words, does the expression become ...

-(x^2-4) < - 3x

Please confirm! Thanks.

Best,
Jay

The inequality in red is the SAME as x² - 4 > 3x.
If we multiply each side of x² - 4 > 3x by -1, we get:
-(x² - 4) < -3x.
Thus, solving -(x² - 4) < -3x is not helpful, since it will yield the same solution set as x² - 4 > 3x.

An easier example:
|x-2| > 5.

Here, there are two cases to consider:
Case 1: The signs in the absolute value are unchanged
x-2 > 5
Case 2: The signs in the absolute value are changed
-(x - 2) > 5
In Case 2, the rest of the inequality remains the same.

Another way!

|x² - 4| > 3x is true for ALL negative values of x. (If x is negative, then the left hand side is positive and the right hand side is negative.) Hence we only need to try positive values of x.

If 2 > x, then |x² - 4| is negative. This gives us |x² - 4| = -(x² - 4) = 4 - x², so we have

4 - x² > 3x, or
0 > x² + 3x - 4, or
0 > (x + 4)(x - 1).

This is true for all x such that -4 < x < 1, so we know that any values between -4 and 1 satisfy the equation. We already knew about the negatives, so this only adds the cases from 0 to 1, not including 1.

Now for the second case! If x > 2, then |x² - 4| is positive. This gives us |x² - 4| = x² - 4, so we have

x² - 4 > 3x, or
x² - 3x - 4 > 0, or
(x - 4)(x + 1) > 0

Remember that in this case we're assuming x > 2, so x is positive. If x is positive, then x must be greater than 4, or (x - 4) is negative. Hence x > 4 also works.

This gives us our range of values: from the first case, x < 1, and from the second, x > 4. Success!

To find our bounds, first look at the equation:
|x^2-4| = 3x
Let's call y = x^2
So |y-4| = 3 SQRT(y)
Square both sides:
y^2 - 8y + 16 = 9y
y^2 - 17y + 16 = 0
(y - 16)(y - 1) = 0
so y = 1, 16
so x = +-1, +-4 (bounds)

Now use arbitrary easy test values between consecutive bounds:
Example,
Less than -4, use x = -10
-4 to -1, use x = -2
-1 to 1, use x = 0
1 to 4, use x = 2
Greater than 4, use x = 10

Substituting these test values into
|x^2-4| > 3x
yields that only x = 2 does not work

Therefore x<1, x>4

Hey guys, I profoundly thank everyone that offered their solutions. It is amazing to see in how many ways a problem could be solved. As much as I love to learn multiple approaches, I tend to confuse myself as I get easily mixed up. So, I stuck to one approach which pretty much worked so far (at least that's my impression of it) until I stumbled on this particular expression in question.

The way I learned to solve a quadratic expression within a modulus operator & with an inequality, is to just solve the expression to arrive at two roots. And then, plot those roots on the number line. Depending on the inequality operator, I would pick those solutions(from the aforementioned roots) that correspond to a specific region. For e.g., if the expression has a < operator, I'd pick the roots that fall in the negative region on the number line and vice versa. For ease of understanding I am re-posting the link with a pictorial representation of what I am trying to explain. Usually, this method gives the solution pretty accurately. For reasons I cannot understand, I am not getting the correct range.

Now, I don't know if "plotting roots on the number line" is conventional approach to solve problems like these, but it seems to be fool-proof to me. Again, I honestly appreciate every solution presented and am in no way discounting them.

Here is the linky - https://s25.postimg.cc/a9mppfx0v/Graph.png (see if this makes any sense to you guys).
Best
Jay.

Yes, plotting roots on a number line is probably the best way.
You could still plot them in the above method. I just visualised them in my head, but I would highly recommend that people draw and write things down. It's only then that errors are easy to see.

Blackboxx wrote:Hey guys, I profoundly thank everyone that offered their solutions. It is amazing to see in how many ways a problem could be solved. As much as I love to learn multiple approaches, I tend to confuse myself as I get easily mixed up. So, I stuck to one approach which pretty much worked so far (at least that's my impression of it) until I stumbled on this particular expression in question.

The way I learned to solve a quadratic expression within a modulus operator & with an inequality, is to just solve the expression to arrive at two roots. And then, plot those roots on the number line. Depending on the inequality operator, I would pick those solutions(from the aforementioned roots) that correspond to a specific region. For e.g., if the expression has a < operator, I'd pick the roots that fall in the negative region on the number line and vice versa. For ease of understanding I am re-posting the link with a pictorial representation of what I am trying to explain. Usually, this method gives the solution pretty accurately. For reasons I cannot understand, I am not getting the correct range.

Now, I don't know if "plotting roots on the number line" is conventional approach to solve problems like these, but it seems to be fool-proof to me. Again, I honestly appreciate every solution presented and am in no way discounting them.

Here is the linky - https://s25.postimg.cc/a9mppfx0v/Graph.png (see if this makes any sense to you guys).
Best
Jay.