Root x +x square=y
1by 2 rootx +2x
Explain this answer
The given equation is root x + x^2 = y. To find the value of x that satisfies this equation, you need to isolate x.
Let's solve the equation step by step:
1. Subtract y from both sides of the equation:
root x + x^2 - y = 0
2. Multiply both sides of the equation by the conjugate of the term containing the square root, which is root x - x^2:
(root x + x^2 - y)(root x - x^2) = 0(root x - x^2)
3. Simplify the left side of the equation using the difference of squares formula:
(root x)^2 - (x^2)^2 = 0
This simplifies to:
x - x^4 = 0
4. Rearrange the terms to put the equation in a more manageable form:
-x^4 + x = 0
5. Factor out an x from the equation:
x(-x^3 + 1) = 0
6. From this equation, we can see that x = 0 is one solution.
Now, we need to solve the equation -x^3 + 1 = 0 for the remaining solutions.
7. Subtract 1 from both sides of the equation:
-x^3 = -1
8. Divide both sides of the equation by -1:
x^3 = 1
9. Take the cube root of both sides:
x = 1
So, the solutions for x that satisfy the equation root x + x^2 = y are x = 0 and x = 1.