Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 25*x^3-10*x^2+x=0
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Equation x^2+7x=18
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Equation (y^4+4*y^2-12)/4=0
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Equation -x=5
- Express {x} in terms of y where:
- -16*x+14*y=-9
- 4*x-7*y=12
- -18*x-14*y=12
- 11*x-2*y=-9
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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
- xx+3x-4=0
- 12x+3(x-4)=-7
- (x-3)*(x-4)=0
- (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