Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 6/x-2+5/x=3
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Equation 4-x/7=x/9
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Equation 0,084:(6,2-x)=1,2
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Equation 3+y-1=0
- Express {x} in terms of y where:
- x^2+y^2=a^2
- x^2+(y-3√x^2)^2=1
- (x^2+y^2)(x+y-3)=2xy
- (-x^2)*y^3+(x^2+y^2-1)^3=0
- Derivative of:
- (x^2+5*x)/(x-4)
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Identical expressions
- (x^ two + five *x)/(x- four)= zero
- (x squared plus 5 multiply by x) divide by (x minus 4) equally 0
- (x to the power of two plus five multiply by x) divide by (x minus four) equally zero
- (x2+5*x)/(x-4)=0
- x2+5*x/x-4=0
- (x²+5*x)/(x-4)=0
- (x to the power of 2+5*x)/(x-4)=0
- (x^2+5x)/(x-4)=0
- (x2+5x)/(x-4)=0
- x2+5x/x-4=0
- x^2+5x/x-4=0
- (x^2+5*x)/(x-4)=O
- (x^2+5*x) divide by (x-4)=0
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Similar expressions
- (x^2-5*x)/(x-4)=0
- (x^2+5*x)/(x+4)=0
(x^2+5*x)/(x-4)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\frac{x^{2} + 5 x}{x - 4} = 0$$
Multiply the equation sides by the denominators:
-4 + x
we get:
$$\frac{\left(x - 4\right) \left(x^{2} + 5 x\right)}{x - 4} = 0$$
$$x \left(x + 5\right) = 0$$
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
$$a = 1$$
$$b = 5$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 0$$
$$x_{2} = -5$$
$$\frac{x^{2} + 5 x}{x - 4} = 0$$
Multiply the equation sides by the denominators:
-4 + x
we get:
$$\frac{\left(x - 4\right) \left(x^{2} + 5 x\right)}{x - 4} = 0$$
$$x \left(x + 5\right) = 0$$
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
$$a = 1$$
$$b = 5$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(5)^2 - 4 * (1) * (0) = 25
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 0$$
$$x_{2} = -5$$