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25x^2-10x+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 25$$
$$b = -10$$
$$c = 1$$
, then
Because D = 0, then the equation has one root.
$$x_{1} = \frac{1}{5}$$
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 25$$
$$b = -10$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(-10)^2 - 4 * (25) * (1) = 0
Because D = 0, then the equation has one root.
x = -b/2a = --10/2/(25)
$$x_{1} = \frac{1}{5}$$
Vieta's Theorem
rewrite the equation
$$\left(25 x^{2} - 10 x\right) + 1 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{2 x}{5} + \frac{1}{25} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{2}{5}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{25}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{2}{5}$$
$$x_{1} x_{2} = \frac{1}{25}$$
$$\left(25 x^{2} - 10 x\right) + 1 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{2 x}{5} + \frac{1}{25} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{2}{5}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{25}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{2}{5}$$
$$x_{1} x_{2} = \frac{1}{25}$$
Sum and product of roots
[src]
sum
1/5
$$\frac{1}{5}$$
=
1/5
$$\frac{1}{5}$$
product
1/5
$$\frac{1}{5}$$
=
1/5
$$\frac{1}{5}$$
1/5
The graph