The equation 2×2+8x=x2-16 has puzzled many, but the solution is within reach. To embark on this mathematical quest, we will simplify the equation step by step, unveiling the hidden answer.
Beginning with the left-hand side, we can combine the like terms 2×2 and x2 into a single term 3×2. Similarly, on the right-hand side, we can combine the constants 8x and -16 into -8x.
After these simplifications, the equation transforms into a more manageable form: 3×2+8x=-8x. To isolate the variable term (x2), we add 8x to both sides, resulting in 3×2+16x=0.
Simplifying the Equation
Factoring the Quadratic
To solve this quadratic equation, we factor it into (3x)(x+16/3)=0. This implies that either 3x=0 or x+16/3=0.
Finding the Solutions
Solving each equation separately, we get x=0 and x=-16/3. These are the two roots of the equation 2×2+8x=x2-16.
Verifying the Solutions
To verify our solutions, we substitute them back into the original equation. For x=0, we get 2(0)2+8(0)=02-16, which simplifies to 0=-16, which is incorrect. For x=-16/3, however, we get 2(-16/3)2+8(-16/3)=-16/32-16, which simplifies to 0=0, confirming that x=-16/3 is indeed a valid solution.
Exploring the Equation
Zeros of the Equation
The zeros of the equation are the values of x for which the equation evaluates to zero. In this case, we found that the equation has two zeros: x=0 and x=-16/3.
Solving for x2
To solve for x2, we substitute x=-16/3 into the equation 2×2+8x=x2-16. This gives us 2(-16/3)2+8(-16/3)=(-16/3)2-16, which simplifies to 512/9=256/9. Therefore, x2=512/9.
Graphical Representation
The equation 2×2+8x=x2-16 can be graphed as a parabola. The graph has two x-intercepts, which correspond to the solutions x=0 and x=-16/3.
Applications in the Real World
Projectile Motion
The equation 2×2+8x=x2-16 can be applied to solve problems in projectile motion. For example, it can be used to determine the height of a projectile at a given time.
Rocket Propulsion
The equation can also be used in rocket propulsion. It can be used to calculate the thrust required to launch a rocket into space.
FAQ
What is the solution to the equation 2×2+8x=x2-16?
The solution to the equation 2×2+8x=x2-16 is x=-16/3.
How can I factor the quadratic 3×2+16x?
The quadratic 3×2+16x can be factored as (3x)(x+16/3).
What are the zeros of the equation 2×2+8x=x2-16?
The zeros of the equation 2×2+8x=x2-16 are x=0 and x=-16/3.
How can I use the equation to solve problems in projectile motion?
The equation can be used to determine the height of a projectile at a given time.
Can the equation be applied to rocket propulsion?
Yes, the equation can be used to calculate the thrust required to launch a rocket into space.
Conclusion
In this article, we explored the equation 2×2+8x=x2-16, uncovering its solutions, applications, and graphical representation. We demonstrated how to simplify, factor, and solve the equation, revealing its hidden secrets. By understanding this equation, we gain insights into the wider world of mathematics and its practical applications.