Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation ((1/5))^(x)^(2)*5^(2*x+2)=(1/5)^(6)
-
Equation x^3-x^2-x-1=0
-
Equation (x+2)^4-4*(x+2)^2-5=0
-
Equation -x=3
- Express {x} in terms of y where:
- 9*x+10*y=5
- -4*x+19*y=-13
- -5*x-6*y=-1
- -7*x+15*y=18
-
Identical expressions
- (5x- three)(x- two)= zero
- (5x minus 3)(x minus 2) equally 0
- (5x minus three)(x minus two) equally zero
- 5x-3x-2=0
- (5x-3)(x-2)=O
-
Similar expressions
- (5x+3)(x-2)=0
- (3-5x)^-(3x-2)(5x+1)
- (5x-3)(x+2)=0
(5x-3)(x-2)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(x - 2\right) \left(5 x - 3\right) = 0$$
We get the quadratic equation
$$5 x^{2} - 13 x + 6 = 0$$
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
$$a = 5$$
$$b = -13$$
$$c = 6$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
$$x_{2} = \frac{3}{5}$$
$$\left(x - 2\right) \left(5 x - 3\right) = 0$$
We get the quadratic equation
$$5 x^{2} - 13 x + 6 = 0$$
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
$$a = 5$$
$$b = -13$$
$$c = 6$$
, then
D = b^2 - 4 * a * c =
(-13)^2 - 4 * (5) * (6) = 49
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
$$x_{2} = \frac{3}{5}$$
Sum and product of roots
[src]
sum
2 + 3/5
$$\frac{3}{5} + 2$$
=
13/5
$$\frac{13}{5}$$
product
2*3 --- 5
$$\frac{2 \cdot 3}{5}$$
=
6/5
$$\frac{6}{5}$$
6/5