What are the real or imaginary solution of the polynomials x^4 -52x^2 +576=0
let u^2=x^4, so you have now
u^2-52u+576=0
(u-26)^2=0
u= 26
x^2=26
u= +- sqrt 26
check that
To find the real or imaginary solutions of the polynomial equation x^4 - 52x^2 + 576 = 0, we can use a simple substitution to transform it into a quadratic equation.
Let's introduce a new variable, y = x^2. Substitute y into the original equation:
y^2 - 52y + 576 = 0
Now we have a quadratic equation in terms of y. We can solve it using factoring, completing the square, or using the quadratic formula. In this case, let's use factoring:
(y - 36)(y - 16) = 0
Setting each factor equal to zero gives us two possible values for y:
y - 36 = 0 or y - 16 = 0
Solving each equation for y, we have:
y = 36 or y = 16
Now, we substitute back for y using y = x^2:
x^2 = 36 or x^2 = 16
Taking the square root of both sides of each equation, we get:
x = ±6 or x = ±4
So the real solutions to the polynomial equation x^4 - 52x^2 + 576 = 0 are x = 6 and x = -6, and the imaginary solutions are x = 4i and x = -4i.
To find the real or imaginary solutions to the polynomial equation x^4 - 52x^2 + 576 = 0, we can use a process called factoring. However, factoring a quartic equation like this is not always straightforward. In this case, we can make use of a substitution to transform the given equation into a quadratic equation, which can then be factored easily.
Let's substitute x^2 = t. When we make this substitution, the equation becomes:
(t)^2 - 52(t) + 576 = 0
Now, we have a quadratic equation in t. To factor this quadratic equation, we can use the fact that the product of the constant term (in this case, 576) and the leading coefficient (in this case, 1) will give us the constant term of the factors (in this case, the a and c values in the factored form). We need to find two numbers whose product is 576 and whose sum is -52 (the coefficient of t). By considering factors of 576, we can find that the numbers -4 and -48 fulfill these requirements.
So, the factored form of the equation now becomes:
(t - 4)(t - 48) = 0
Now, we can solve for t by setting each factor equal to zero:
t - 4 = 0 or t - 48 = 0
Solving these equations will give us the values of t:
t = 4 or t = 48
Since we initially substituted x^2 = t, we can substitute these values back to get the solutions for x:
For t = 4:
x^2 = 4
Taking the square root of both sides:
x = ±2
For t = 48:
x^2 = 48
Taking the square root of both sides:
x = ±√48 = ±4√3
Therefore, the real solutions to the polynomial equation x^4 - 52x^2 + 576 = 0 are x = -2, x = 2, x = -4√3, and x = 4√3.