The following equation has three solutions. Look for the greatest common factor, then use the quadratic formula to find all solutions.
10t^5 + 5t^4 = 50t^3
10t^5 + 5t^4 - 50t^3 = 0
t^3( 2t^2 + t - 10) = 0
t^3(2t+5)(t-2) = 0
t = 0 or t = -5/2 or t = 2
To find the solutions to the equation 10t^5 + 5t^4 = 50t^3, we first need to rearrange the equation to set it equal to zero:
10t^5 + 5t^4 - 50t^3 = 0
Next, we can factor out the greatest common factor from each term. In this case, the common factor is t^3:
t^3(10t^2 + 5t - 50) = 0
Now we have factored out the greatest common factor, we can focus on solving the quadratic equation within the parentheses. To solve the quadratic equation, we can use the quadratic formula:
For an equation of the form ax^2 + bx + c = 0, the quadratic formula states that the solutions for x are:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation 10t^2 + 5t - 50 = 0, a = 10, b = 5, and c = -50. We can substitute these values into the quadratic formula:
t = (-5 ± √(5^2 - 4 * 10 * -50)) / (2 * 10)
Simplifying further, we have:
t = (-5 ± √(25 + 2000)) / 20
t = (-5 ± √2025) / 20
t = (-5 ± 45) / 20
This gives us two solutions:
t = (-5 + 45) / 20 = 40 / 20 = 2
t = (-5 - 45) / 20 = -50 / 20 = -2.5
Finally, we also have the solution t^3 = 0, which means t = 0.
Therefore, the three solutions to the equation 10t^5 + 5t^4 = 50t^3 are t = 2, t = -2.5, and t = 0.