Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2-2x=0
-
Equation x^2-3x-10=0
-
Equation (x-119)/(x+7)=-5
-
Equation x/4+13=x+1
- Express {x} in terms of y where:
- 15*x-5*y=-5
- 18*x+20*y=-2
- 15*x+7*y=2
- 12*x-20*y=10
-
Identical expressions
- (x+ three)*(x- four)= zero
- (x plus 3) multiply by (x minus 4) equally 0
- (x plus three) multiply by (x minus four) equally zero
- (x+3)(x-4)=0
- x+3x-4=0
- (x+3)*(x-4)=O
-
Similar expressions
- -4*(5-2x)+3*(x-4)=(6*2-x)-5x
- (x+3)(x-4)-(x-7)(x-2)=2
- 12x+3(x-4)=-7
- (x-3)*(x-4)=0
- (-10*x*x+3*x-40)^2=2*(x*x+1)*(100*x*x-60*x+16)
- (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
$$x^{2} - x - 12 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -12$$
, 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
$$x^{2} - x - 12 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (-12) = 49
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 4$$
$$x_{2} = -3$$
Sum and product of roots
[src]
sum
-3 + 4
$$-3 + 4$$
=
1
$$1$$
product
-3*4
$$- 12$$
=
-12
$$-12$$
-12
The graph