Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation sin(x)=4
-
Equation 5^x=125
-
Equation (|x+4|)=5
-
Equation 2*sin(x)=1
- Express {x} in terms of y where:
- 16*x-17*y=-15
- -4*x-4*y=19
- 15*x+1*y=19
- -5*x-2*y=9
-
Identical expressions
- (x+ three)*(x- two)= zero
- (x plus 3) multiply by (x minus 2) equally 0
- (x plus three) multiply by (x minus two) equally zero
- (x+3)(x-2)=0
- x+3x-2=0
- (x+3)*(x-2)=O
-
Similar expressions
- (x+3)*(x+2)=0
- 14*x-9/17*x+3*x-25/17*x=0
- (x-3)*(x-2)=0
- (2*x+3)*(x-2)^2-(1-14*x)*(x^2-4)+54*x^2+108*x=0
- (x+3)(x-2)-(x+4)(x-1)=3x
(x+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
$$\left(x - 2\right) \left(x + 3\right) = 0$$
We get the quadratic equation
$$x^{2} + x - 6 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -6$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
$$x_{2} = -3$$
$$\left(x - 2\right) \left(x + 3\right) = 0$$
We get the quadratic equation
$$x^{2} + x - 6 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -6$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (-6) = 25
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
$$x_{2} = -3$$
Sum and product of roots
[src]
sum
-3 + 2
$$-3 + 2$$
=
-1
$$-1$$
product
-3*2
$$- 6$$
=
-6
$$-6$$
-6
The graph