Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x+5=7
-
Equation 5x^2-8x+3=0
-
Equation 3/5*y=29/10-1/5
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Equation -12x+11=-(2-x)
- Express {x} in terms of y where:
- -11*x+16*y=3
- -4*x-13*y=-14
- 19*x-1*y=2
- -2*x+20*y=-3
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Identical expressions
- (x- one)*(x- four)
- (x minus 1) multiply by (x minus 4)
- (x minus one) multiply by (x minus four)
- (x-1)(x-4)
- x-1x-4
-
Similar expressions
- (x+1)*(x-4)
- (-x-1)*(x-4)=x*(4*x+3)
- (x-4)(x-1)+5(x-1)(x-4)=0
- (x-1)(x-4)+2=0
- (x+8)2-(2x-1)(x-4)=60
- (x-3)/((x-1)(x-4))+((x-1)(x-4))/(x-3)=(5)/(2)
- (x-1)*(x+4)
(x-1)*(x-4) equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(x - 4\right) \left(x - 1\right) = 0$$
We get the quadratic equation
$$x^{2} - 5 x + 4 = 0$$
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
$$a = 1$$
$$b = -5$$
$$c = 4$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 4$$
$$x_{2} = 1$$
$$\left(x - 4\right) \left(x - 1\right) = 0$$
We get the quadratic equation
$$x^{2} - 5 x + 4 = 0$$
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
$$a = 1$$
$$b = -5$$
$$c = 4$$
, then
D = b^2 - 4 * a * c =
(-5)^2 - 4 * (1) * (4) = 9
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 4$$
$$x_{2} = 1$$