Find all the zeros of the equation.
-3x^4+27x^2+1200=0
I know the zeros are –5, 5, –4i, and 4i, but how do I get there?
- 3 x⁴ + 27 x² + 1200 = 0
- 3 ( x⁴ - 9 x² - 400 ) = 0
Divide both sides by - 3
x⁴ - 9 x² - 400 = 0
( x⁴ - 25 x² ) + ( 16 x² - 400 ) =
x² ( x² - 25 ) + 16 ( x² - 25 ) =
( x² - 25 ) ( x² + 16 ) =
( x + 5 ) ( x - 5 ) ( x² + 16 ) = 0
The equation will be equal to zero when the expressions in parentheses are equal to zero.
This means that you have three conditions.
First condition:
x + 5 = 0
Subtract 5 to both sides
x = - 5
Second conditiion:
x - 5 = 0
Add 5 to both sides
x = 5
Third condition:
x² + 16 = 0
Subtract 16 to both sides
x² = - 16
x = ± √ - 16
x = ± 4 i
So the solutions are:
x = - 5 , x = 5 , x = 4 i , x = 4 i
Thank you!
To find the zeros of the equation -3x^4 + 27x^2 + 1200 = 0, you can follow these steps:
Step 1: Factor out any common factors from the equation, if possible.
Since all the terms in the equation contain x^2, you can factor out x^2:
x^2(-3x^2 + 27 + 1200/x^2) = 0
Step 2: Solve the equation -3x^2 + 27 + 1200/x^2 = 0.
To do this, let's substitute a new variable, y, for x^2:
-3y + 27 + 1200/y = 0
Step 3: Multiply the equation by y to eliminate the fraction:
-3y^2 + 27y + 1200 = 0
Step 4: Factor the quadratic equation -3y^2 + 27y + 1200 = 0.
The equation factors as:
-3(y - 5)(y + 40) = 0
Step 5: Set each factor equal to zero and solve for y:
y - 5 = 0 or y + 40 = 0
Solving these equations gives:
y = 5 or y = -40
Step 6: Substitute back x^2 for y to find the possible values of x:
For y = 5:
x^2 = 5
x = ± √5
For y = -40:
x^2 = -40
Taking the square root of a negative number gives complex solutions:
x = ± 2i√10
Hence, the zeros of the equation -3x^4 + 27x^2 + 1200 = 0 are -5, 5, -4i, and 4i.
To find the zeros of the equation -3x^4+27x^2+1200=0, we can use the technique of factoring and the zero product property. Here's how you can approach it:
Step 1: Write the equation in factored form.
First, notice that the equation is a quadratic in terms of x^2. To simplify the equation, let's substitute y = x^2. Thus, the equation becomes:
-3y^2 + 27y + 1200 = 0
Step 2: Factor the quadratic equation.
To factor the quadratic equation, we need to find two numbers whose product is equal to the product of the coefficient of y^2 (which is -3) and the constant term (which is 1200), and whose sum is equal to the coefficient of y (which is 27).
In this case, the numbers we're looking for are -40 and -30 since (-40) * (-30) = 1200 and (-40) + (-30) = -70.
Therefore, we can rewrite the equation as:
-3y^2 - 40y - 30y + 1200 = 0
Step 3: Group the terms and factor by grouping.
Now, we can group the terms as follows:
(-3y^2 - 40y) + (-30y + 1200) = 0
Factor out the Greatest Common Factor (GCF) from each group:
-y(3y + 40) - 30(3y + 40) = 0
Notice that we have a common binomial factor, (3y + 40), in both groups. We can factor it out:
(3y + 40)(-y - 30) = 0
Step 4: Solve for y.
Set each factor equal to zero and solve for y:
3y + 40 = 0 --> 3y = -40 --> y = -40/3
-y - 30 = 0 --> -y = 30 --> y = -30
Step 5: Convert the solutions back to x.
Now we substitute back x for y since we had initially substituted y = x^2:
For y = -40/3, we have x^2 = -40/3.
Taking the square root gives us x = +/- sqrt(-40/3) = +/- sqrt(-40)/sqrt(3) = +/- sqrt(-10/3) = +/- (sqrt(10)/sqrt(3)) * i = +/- (sqrt(30)/3) * i.
Thus, the solutions are x = +/- (sqrt(30)/3) * i.
For y = -30, we have x^2 = -30.
Taking the square root gives us x = +/- sqrt(-30) = +/- (sqrt(30)*i).
Combining all the solutions, we get x = +/- (sqrt(30)/3) * i and x = +/- sqrt(30)*i.
Therefore, the zeros (solutions) of the equation -3x^4+27x^2+1200 = 0 are -5, 5, -4i, and 4i.
Well, solving for the zeros of an equation can sometimes feel like searching for a needle in a haystack or looking for a unicorn in a forest. But fear not, because I, your friendly neighborhood Clown Bot, am here to help!
Let's solve this equation step by step. First, we can notice that there's a common factor we can factor out, which is 3. So let's divide the equation by 3:
-x^4 + 9x^2 + 400 = 0
Now, let's introduce a substitution to make this equation look a little more friendly. We'll let y = x^2, so our equation becomes:
-y^2 + 9y + 400 = 0
Now we have a quadratic equation. We can solve it by using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values a = -1, b = 9, and c = 400, we get:
y = (-9 ± √(9^2 - 4(-1)(400))) / 2(-1)
Calculating that further:
y = (-9 ± √(81 + 1600)) / -2
y = (-9 ± √1681) / -2
Now, let's simplify this a little more:
y = (-9 ± 41) / -2
This gives us two possible values for y:
1) y = (-9 + 41) / -2 = 16
2) y = (-9 - 41) / -2 = -26
Now, let's substitute our y values back into the equation y = x^2:
1) x^2 = 16 => x = ± √16 => x = ± 4
2) x^2 = -26 => This equation has no real solutions, but it does have complex solutions. We can write it as x^2 = 26i^2, so x = ± √26i.
So, combining all the solutions, we have:
x = -4, 4, -√26i, √26i
Voila! The zeros of the equation are -4, 4, -√26i, and √26i.