Question

25y^3 - 10y^2 + y = 0

Answer 1

y(25y^2-10y+1)=0

y1=0

25y^2-10y+1=0

D= b^2-4ac = 100 - 4*25*1 = 100 - 100 = 0

y2= -b+√D/2a = 10+0/50 = 1/5

Answer 2

To solve the equation 25y^3 - 10y^2 + y = 0, we need to find the values of y that satisfy the equation. This is a cubic equation, so it may have up to three real solutions.

Step 1: Factor out the common term 'y' from the equation: y(25y^2 - 10y + 1) = 0

Step 2: Solve the quadratic equation 25y^2 - 10y + 1 = 0. We can use the quadratic formula to find the roots of this equation.

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 25, b = -10, and c = 1.

Step 3: Calculate the discriminant (D) to determine the nature of the roots. D = b^2 - 4ac D = (-10)^2 - 4 * 25 * 1 D = 100 - 100 D = 0

Since the discriminant is equal to zero, the quadratic equation has two identical real roots.

Step 4: Use the quadratic formula to find the roots of the quadratic equation. y = (-(-10) ± √(0)) / (2 * 25) y = (10 ± 0) / 50

So the two identical real roots are: y = 10 / 50 = 1/5

Step 5: Consider the factor 'y' from Step 1. y = 0

The solutions to the original cubic equation are: y = 1/5, 0