What are the real or imaginary solutions of the polynomial equation? x4 – 52x2 + 576 = 0 (1 point) Responses 4, –4 4, –4 4, –6 4, –6 4, –4, 6, –6 4, –4, 6, –6 no solution
The real solutions of the equation can be found by factoring the equation:
x^4 - 52x^2 + 576 = 0.
We can rewrite the equation as:
(x^2 - 16)(x^2 - 36) = 0.
Setting each factor equal to zero:
x^2 - 16 = 0,
(x - 4)(x + 4) = 0,
x = 4, -4.
x^2 - 36 = 0,
(x - 6)(x + 6) = 0,
x = 6, -6.
So the real solutions of the equation are x = 4, -4, 6, -6.
Therefore, the correct response is:
4, -4, 6, -6.
To find the solutions of the polynomial equation x^4 - 52x^2 + 576 = 0, we can use factoring or the quadratic formula.
Let's use a substitution to simplify the equation:
Let y = x^2
Substituting this in the equation gives us:
y^2 - 52y + 576 = 0
Now we can factor this quadratic equation:
(y - 16)(y - 36) = 0
Setting each factor equal to zero:
y - 16 = 0 or y - 36 = 0
Solving for y:
y = 16 or y = 36
Now we substitute back the value of y into the equation y = x^2:
x^2 = 16 or x^2 = 36
Taking square roots of both sides:
x = ±√16 or x = ±√36
So the solutions are:
x = ±4 or x = ±6
Therefore, the real or imaginary solutions of the polynomial equation x^4 - 52x^2 + 576 = 0 are:
x = ±4 and x = ±6.
To find the real or imaginary solutions of the polynomial equation x^4 – 52x^2 + 576 = 0, we need to solve for the values of x that satisfy the equation.
First, let's rewrite the equation in a quadratic form. We can substitute a variable, let's say y = x^2, to obtain a quadratic equation:
y^2 – 52y + 576 = 0.
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula, which states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = -52, and c = 576.
Using the quadratic formula, we can calculate the solutions for y:
y = (-(-52) ± √((-52)^2 - 4(1)(576))) / (2(1))
y = (52 ± √(2704 - 2304)) / 2
y = (52 ± √400) / 2
y = (52 ± 20) / 2
Now, we have two possibilities:
1) y = (52 + 20) / 2 = 72/2 = 36
In this case, we can solve for x: x^2 = 36.
Taking the square root of both sides, we get x = ±6.
2) y = (52 - 20) / 2 = 32/2 = 16
In this case, we can solve for x: x^2 = 16.
Taking the square root of both sides, we get x = ±4.
So, the solutions for the original equation x^4 – 52x^2 + 576 = 0 are x = 6, -6, 4, -4.
Therefore, the correct response is: 4, -4.