Solve the polynomial equation by finding all real roots
x^4+15x^2-16
Many thanks
let a = x^2 ... a^2 + 15 a - 16 ... (a + 16) (a - 1)
(x^2 + 16) (x^2 - 1) ... (x + 4i) (x - 4i) (x + 1) (x - 1)
To solve the polynomial equation x^4 + 15x^2 - 16 = 0 and find all real roots, we can use a technique called factoring.
Step 1: Rewrite the equation as a quadratic equation in terms of x^2:
Let's substitute a variable, say u, as x^2. Then the equation becomes u^2 + 15u - 16 = 0.
Step 2: Factor the quadratic equation:
To factor u^2 + 15u - 16, we need to find two numbers that when multiplied give -16 and when added give 15. The numbers that satisfy these conditions are 16 and -1. So we can rewrite the equation as (u + 16)(u - 1) = 0.
Step 3: Solve for u:
Setting each factor equal to zero, we have u + 16 = 0 and u - 1 = 0. Solving for u, we get u = -16 and u = 1.
Step 4: Solve for x:
Remember that u is x^2, so substituting back, we have x^2 = -16 and x^2 = 1.
For x^2 = -16:
Taking the square root of both sides, we get x = √(-16). Since we are looking for real roots, there are no solutions for x in this case.
For x^2 = 1:
Taking the square root of both sides, we get x = ±√1. So x = 1 and x = -1 are the real roots of the polynomial equation.
Therefore, the real roots of the polynomial equation x^4 + 15x^2 - 16 = 0 are x = 1 and x = -1.