The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.
In this example, it was relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.
We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.
But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:
Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0
In this problem, x can be any real number – we have no constraints on it. Now, is x negative?
Statement 1: x^3 + x^2 + x + 2 = 0
If we try to solve this equation as we are used to doing, look at what happens:
If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0
We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.
Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.
Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.
This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.
Statement 2: x^2 – x – 2 < 0
This, we can easily solve:
x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0
We know how to solve this inequality (as discussed in our Inequality module here)
This will give us -1 < x < 2.
Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.
To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.
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