Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 4-6*(x+2)=3-5*x
-
Equation 1/(x-3)^2-3/(x-3)-4=0
-
Equation x^2-2x+1=0
-
Equation 3(x+3)=2x+5
- Express {x} in terms of y where:
- 2*x+13*y=19
- 17*x-20*y=6
- 5*x-13*y=17
- -9*x-1*y=6
-
Identical expressions
- (-x- four)(three x+3)= zero
- ( minus x minus 4)(3x plus 3) equally 0
- ( minus x minus four)(three x plus 3) equally zero
- -x-43x+3=0
- (-x-4)(3x+3)=O
-
Similar expressions
- (-x-4)*(3*x+3)=0
- (-x+4)(3x+3)=0
- (x-4)(3x+3)=0
- (-x-4)(3x-3)=0
- (-x-4)*(3x+3)=0
(-x-4)(3x+3)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(- x - 4\right) \left(3 x + 3\right) = 0$$
We get the quadratic equation
$$- 3 x^{2} - 15 x - 12 = 0$$
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
$$a = -3$$
$$b = -15$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -4$$
$$x_{2} = -1$$
$$\left(- x - 4\right) \left(3 x + 3\right) = 0$$
We get the quadratic equation
$$- 3 x^{2} - 15 x - 12 = 0$$
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
$$a = -3$$
$$b = -15$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(-15)^2 - 4 * (-3) * (-12) = 81
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -4$$
$$x_{2} = -1$$
Sum and product of roots
[src]
sum
-4 - 1
$$-4 - 1$$
=
-5
$$-5$$
product
-4*(-1)
$$- -4$$
=
4
$$4$$
4
The graph