Question

X^2+(1-5a)x+4a^2-a=0

Answer 1

$x^2+(1-5a)x+4a^2-a=0 \\ D=(1-5a)^2-4\cdot(4a^2-a)=9a^2-6a+1=(1-3a)^2 \\ x_1= \frac{5a-1+1-3a}{2} = a \\ x_2= \frac{5a-1-1+3a}{2}=4a-1$

Answer 2

The equation you've provided is a quadratic equation in the variable x, with the coefficients involving the parameter 'a'. To solve this quadratic equation for x, you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In your equation, the coefficients are as follows: a = 1 b = 1 - 5a c = 4a² - a

Now, plug these values into the quadratic formula:

x = (-(1 - 5a) ± √((1 - 5a)² - 4(1)(4a² - a))) / (2(1))

x = (-1 + 5a ± √((1 - 5a)² - 4(4a² - a))) / 2

x = (-1 + 5a ± √(1 - 10a + 25a² - 16a² + 4a)) / 2

x = (-1 + 5a ± √(25a² - 16a² - 10a + 4a + 1)) / 2

x = (-1 + 5a ± √(9a² - 6a + 1)) / 2

Now, you can simplify further if necessary or leave it in this form. The solutions for x will be given by:

x₁ = (-1 + 5a + √(9a² - 6a + 1)) / 2 x₂ = (-1 + 5a - √(9a² - 6a + 1)) / 2

These are the solutions to the quadratic equation for the variable x in terms of the parameter 'a'.