Mister Exam
Other calculators
-
How to use it?
- Equation:
- Equation x+y=5
-
Equation tan(x)=-1/2
-
Equation 1*(x-1)*(x-3)=0
-
Equation 5x+12=8x+30
- Express {x} in terms of y where:
- -2*x-19*y=3
- 5*x+4*y=-12
- -18*x-14*y=12
- -9*x+1*y=3
-
Identical expressions
- one *(x- one)*(x- three)= zero
- 1 multiply by (x minus 1) multiply by (x minus 3) equally 0
- one multiply by (x minus one) multiply by (x minus three) equally zero
- 1(x-1)(x-3)=0
- 1x-1x-3=0
- 1*(x-1)*(x-3)=O
-
Similar expressions
- 1*(x-1)*(x+3)=0
- (x+2)(x+1)(x-1)(x-3)=0
- 1*(x+1)*(x-3)=0
1*(x-1)*(x-3)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(x - 3\right) \left(x - 1\right) = 0$$
We get the quadratic equation
$$x^{2} - 4 x + 3 = 0$$
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
$$a = 1$$
$$b = -4$$
$$c = 3$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 3$$
$$x_{2} = 1$$
$$\left(x - 3\right) \left(x - 1\right) = 0$$
We get the quadratic equation
$$x^{2} - 4 x + 3 = 0$$
This equation is of the form
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
$$a = 1$$
$$b = -4$$
$$c = 3$$
, then
D = b^2 - 4 * a * c =
(-4)^2 - 4 * (1) * (3) = 4
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 3$$
$$x_{2} = 1$$
The graph