Factor and solve for x
1) x^4 - 9x^2> 0
2) 4x2^x - x^3 2^x<or equal to 0
I don't know where to start on 1 but for 2 I got
X<or equal to -4 or x> or equal to 4
Could some one please explain how to do 1 and tell me if I did 2 correctly?
1)x^2(x^2-9)=x^2(x-3)(x+3)>0
x>3 or x<-3
2)x*2^x(4-x^2)=x*2^x(2-x)(2+x)<=0
-2<=x<=0 or x>=2
How did you get those 2 answers for 2?
a^2-b^2=(a-b)(a+b)
4-x^2=2^2-x^2
To solve these equations, we need to factor the expressions and find the values of x that satisfy the given inequalities.
1) x^4 - 9x^2 > 0:
First, notice that this equation can be factored by grouping:
x^4 - 9x^2 = (x^2 - 3x)(x^2 + 3x)
Now we have two factors: (x^2 - 3x) and (x^2 + 3x). We need to determine when the product of these factors is greater than zero.
To do this, we examine the signs of each factor separately:
(a) x^2 - 3x > 0:
We can factor x out of this equation to obtain: x(x - 3) > 0.
This equation holds true if both x and (x - 3) have the same sign.
Considering the sign of x - 3:
- If x - 3 > 0, or x > 3, then both x and (x - 3) are positive.
- If x - 3 < 0, or x < 3, then both x and (x - 3) are negative.
(b) x^2 + 3x > 0:
This quadratic equation does not factor further easily, so we will use another method.
The discriminant is given by b^2 - 4ac, where a = 1, b = 3, and c = 0 for this equation.
The discriminant is 3^2 - 4(1)(0) = 9, which is positive.
Since the discriminant is positive, this means that the quadratic has two distinct real roots. Thus, the signs will change at these roots.
Next, we find the roots by setting x^2 + 3x = 0:
x(x + 3) = 0
Either x = 0 or x = -3.
We can create an interval chart with these critical points:
---------+------0-------+--------3-------+---------
x | -3 | 0 | 3
---------+------|-------+--------|-------+---------
(x^2-3x) | - | + | - |
(x^2+3x) | - | - | + |
---------+------|-------+--------|-------+---------
Reading from left to right, we see that x^2 - 3x is positive when x < 0 or x > 3.
Similarly, x^2 + 3x is positive when -3 < x < 0.
To satisfy the original inequality, x^4 - 9x^2 > 0, both factors must have the same sign (positive). Since they have opposite signs when -3 < x < 0, this portion does not satisfy the inequality.
Finally, the solution to x^4 - 9x^2 > 0 is x < 0 or x > 3.
2) 4x2^x - x^3 2^x ≤ 0:
To solve this inequality, we can begin by factoring the expression:
x^2^x - x^3 * 2^x = x^2^x - 2^x * x^3 = x^2^x(1 - 2^x * x).
Now, let's analyze the factors:
(a) x^2^x = x^(2x)
(b) (1 - 2^x * x): This factor does not factorize easily.
Next, let's consider each factor separately:
(a) x^(2x) > 0:
The exponent is even, so x^(2x) will always be positive (except when x = 0, which doesn't satisfy the given inequality).
(b) (1 - 2^x * x) ≤ 0:
To determine the sign of this factor, we need to examine its behavior for various values of x.
- When x < 0, both factors (1 - 2^x * x) and x^(2x) are positive.
- When x = 0, the expression is undefined due to division by zero.
- When x > 0, the sign of (1 - 2^x * x) will depend on the relative values of 1 and 2^x * x.
From observation, we find that when x > 4, the factor (1 - 2^x * x) is negative. However, the original inequality is not ≤ when x > 4; it is simply <. Therefore, we disregard x > 4 as a solution.
Combining the results from both factors, we see that:
- For x < 0, the inequality is satisfied: x < 0.
- For x = 0, the inequality is not satisfied: x ≠ 0.
- For 0 < x ≤ 4, the inequality is not satisfied: x ≠ 0.
- For x > 4, the inequality is not satisfied: x > 4.
Hence, the solution to 4x2^x - x^3 2^x ≤ 0 is x < 0 or x > 4. You correctly obtained this result.