How to solve x^2>a^2
has to equal x>a but apparently also x<-a
How do I prove that?
First, solve x^2=a^2. Take the plus and minus square root of both sides. The negative root is needed because when that is squared on both sides, you still get back the original equation. ==> x=a, x= -a.
Then, plot this on a horizontal number line. Our domain is ( -∞, ∞), so the line has arrows on both sides to indicate that (if the notation is not familiar, read up on interval notation). If our domain is limited, then the line would instead end in a filled dot or open hole depending on whether that particular end is open or closed. On the horizontal line, draw vertical tick marks. These marks show where x = -a (the left tick mark) and x= +a (the right tick mark). Below the horizontal line, think of the line representing your domain. Above, it represents the range.
To the left of the left tick mark is all x< -a. In between the tick marks is all x greater than -a but less than +a. To the right of the right tick mark is all x > +a. Above the horizontal line, you check if each of the three sections as well as the tick marks satisfies x^2>a^2 (x = -a; x = +a; x < -a; -a < x < +a; x> +a).
For the intervals, all you need to do is check one number within each interval to see if the whole interval satisfies the original equation. Why? It is easier to see if we modify the original equation. Subtract a^2 from both sides, which does not change the direction of the inequality symbol, to get x^2 - a^2 >0. By difference of squares ==> (x+a)(x-a)>0. Essentially, we need to find when the parabola y=(x+a)(x-a) is positive, or greater than zero. We draw the same number line with the same tick marks. Now, when we check each section, that is, our intervals separated by the tick marks, we're checking if that section is positive or negative. The tick marks can only be zero or undefined. Our tick marks only represent when y=0, because the parabola is not undefined for any y value. The section cannot switch signs, or else there would be another tick mark dividing it into two more separate sections (because the parabola is continuous).
Returning to the original approach, neither tick marks satisfies the equation because a^2 is not greater than a^2.
How about the left section (x < -a)? Check x = -a - 1. Then,
x^2 = (-a - 1)^2
x^2 = ((-1)*(a+1))^2 Factor out -1
x^2 = 1*(a+1)^2 Applied power of 2 to -1 and (a+1). (-1)^2 = 1.
x^2 = a^2 +2a +1 Binomial expansion
a^2 + 2a +1 > a^2 Substitute x^2 back into original equation.
2a + 1 > 0 ? Subtract a^2 from both sides. Assuming a > 0 (positive), then the equation holds and holds for all x < -a. Put a check mark above that section on the number line to show that this interval satisfies the original equation.
The middle section. Check x = 0, because -a < 0 < +a.
0 > a^2 ? x^2=0. Substitute that into original equation.
This is false because we assumed a > 0. It would also be false if a < 0 (negative), because a^2 would still be a positive number. All positive numbers are NOT less than zero.Therefore, the middle section does not satisfy the original equation. Put an "X" above that section on the number line to show that this interval does not satisfy the original equation
The section on the right. Check x = a+1. Then x^2 = (a+1)^2 = a^2 + 2a +1. Substitute into original equation.
==> a^2 + 2a +1 > a^2
2a + 1 > 0 ? Subtract a^2 from both sides.
Following the same reasoning as that when analyzing the interval x < -a, we know that x > +a also satisfies the original equation. Put a check mark above the section representing x > +a.
Finally, use interval notation for our answer. All x in ( -∞, -a) ∪ (a,∞) satisfies our original equation. This is equivalent to all x < -a or x>a.
1. Solve x^2 = a^2.
2. Draw horizontal number line. Draw vertical tick marks on the horizontal line showing x-values satisfying x^2 = a^2.
3. Check each interval and tick mark to see if it satisfies x^2 > a^2. For intervals, set an x equal to a value within that interval. Then plug it back into x^2 > a^2.
4a. If that value in the interval satisfies the original equation, then the whole interval satisfies the original equation. An interval can be to the left, to the right, or in between tick marks.
4b. If that value in the interval does not satisfy the original equation, then the whole interval does not satisfy the original equation.
5. Put all the intervals and endpoints/tick marks together for your final answer, preferably in interval notation.
Post is closed for comments.