Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 1+sin(x)=0
-
Equation tg(x)=1
-
Equation x^4+4=0
-
Equation -x-4+5(x+3)=5(-1-x)-2
- Express {x} in terms of y where:
- 4*x+2*y=5
- 1*x+10*y=-18
- -18*x-2*y=9
- -11*x+9*y=-20
-
Identical expressions
- (x- three)*(x- four)= zero
- (x minus 3) multiply by (x minus 4) equally 0
- (x minus three) multiply by (x minus four) equally zero
- (x-3)(x-4)=0
- x-3x-4=0
- (x-3)*(x-4)=O
-
Similar expressions
- (x+3)*(x-4)=0
- (x^2-7x+13)^2-(x-3)(x-4)=1
- (x-3)(x-4)(x-5)=(x-2)(x-4)(x-5)
- 6x-8=5x-3(x-4)
- (x-3)*(x+4)=0
(x-3)*(x-4)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(x - 4\right) \left(x - 3\right) = 0$$
We get the quadratic equation
$$x^{2} - 7 x + 12 = 0$$
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
$$a = 1$$
$$b = -7$$
$$c = 12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 4$$
$$x_{2} = 3$$
$$\left(x - 4\right) \left(x - 3\right) = 0$$
We get the quadratic equation
$$x^{2} - 7 x + 12 = 0$$
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
$$a = 1$$
$$b = -7$$
$$c = 12$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (1) * (12) = 1
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 4$$
$$x_{2} = 3$$
Sum and product of roots
[src]
sum
3 + 4
$$3 + 4$$
=
7
$$7$$
product
3*4
$$3 \cdot 4$$
=
12
$$12$$
12
The graph