%5E2%2B(x%5E2%2B2x-15)%5E2%3D0.jpeg "(x^2-25)^2+(x^2+2x-15)^2=0")
Solving the Equation: (x^2-25)^2+(x^2+2x-15)^2=0
In this article, we will explore the solution to the equation (x^2-25)^2+(x^2+2x-15)^2=0.
Factoring the Equation
Let's start by factoring the left-hand side of the equation:
(x^2-25)^2+(x^2+2x-15)^2=0
We can rewrite the equation as:
(x-5)^4(x+5)^4+(x-3)^2(x+5)^2(x-5)^2=0
Simplifying the Equation
Now, let's simplify the equation by canceling out the common factors:
(x-5)^2(x+5)^2((x-5)^2+(x-3)^2)=0
Finding the Roots
To find the roots of the equation, we set each factor equal to zero and solve for x:
x-5=0 => x=5
x+5=0 => x=-5
x-3=0 => x=3
Since the equation is of the form (x-a)^2(x-b)^2=0, we know that the roots are repeated. Therefore, the roots of the equation are x=5 (with multiplicity 2) and x=-5 (with multiplicity 2).
Conclusion
In conclusion, the roots of the equation (x^2-25)^2+(x^2+2x-15)^2=0 are x=5 and x=-5, each with multiplicity 2.
